第一版前言

Preface to the First Edition

致学生们

To students

你好！

Hello!

您即将开始学习一个令人着迷且重要的领域：计算理论。它涵盖了计算机硬件、软件及其某些应用的基本数学特性。当我们研究这个领域时，我们试图确定什么可以计算，什么不能计算，可以多快完成，需要多少内存，以及采用什么类型的计算模型。这些问题显然与工程实践相关，同时，与许多科学领域一样，它们也具有纯粹的哲学方面。

You are about to begin exploring a fascinating and important field: computational theory. It encompasses the fundamental mathematical properties of computer hardware, software, and some of their applications. In studying this field, we try to determine what can and cannot be computed, how fast it can be done, how much memory is needed, and in what kind of computational model. These issues are obviously related to engineering practice, and at the same time, as in many fields of science, they also have purely philosophical aspects.

我知道你们中的许多人都期待着学习这些材料，尽管也许并非所有人都愿意来到这里。也许您想获得计算机科学或工程学文凭，并且理论讲座就在必需的清单上 - 只有上帝知道为什么。难道理论不是模糊的、无聊的，更糟糕​​的是，无关紧要的吗？

I know that many of you are eager to learn this material, although perhaps not all of you are here willingly. Perhaps you are looking to get a degree in computer science or engineering, and a theory course is on the required list—God knows why. Isn't theory muddy, boring, and worse, irrelevant?

继续读下去，你会发现这个理论既不晦涩难懂，也不乏味，而是很容易理解，甚至很有趣。理论计算机科学包含许多精彩的大想法，但也有一些小而有时无聊的细节，可能会令人厌倦。学习新事物总是很困难，但如果材料呈现得当，它会变得更容易、更有趣。我写这本书的主要目标是向读者展示计算理论的迷人方面，而不是陷入不必要的复杂性。当然，确定你是否会觉得这个理论有趣的唯一方法是尝试学习它。

Read on and you will see that the theory is neither obscure nor arcane nor tedious, but quite understandable and downright interesting. Theoretical computer science contains many wonderful, grand ideas, although it is not without its small and sometimes boring details that can be tiresome. Learning new things is always difficult, but it can be made easier and more enjoyable if the material is properly presented. My main goal in writing this book is to introduce readers to the fascinating aspects of the theory of computation without getting bogged down in unnecessary complexities. Of course, the only way to find out whether you will find this theory interesting is to try to learn it.

理论与实践密切相关。它提供了从业者在计算机工程中使用的思维工具。您是否正在为特殊应用程序设计一种新的编程语言？你学到的语法知识将会派上用场。您进行单词搜索和模式匹配吗？记住有限自动机和正则表达式。您是否面临着一个似乎需要比您分配的时间更多的时间来解决的问题？想想你对 NP 完备性的了解。各种应用领域，例如现代密码协议，都是基于您将在本书中学到的理论原理。

Theory is closely related to practice. It provides the thinking tools that practitioners use in computer engineering. Are you designing a new programming language for a special purpose? What you learn about grammars will come in handy . Are you doing word searches and pattern matching? Remember finite automata and regular expressions . Are you faced with a problem that seems to take longer to solve than you can spare? Think about what you learned about NP-completeness. Various application areas, such as modern cryptographic protocols, rely on the theoretical principles you learn in this book.

该理论也很重要，因为它展示了计算机新的、更简单的和更优雅的一面，而计算机通常被认为是非常复杂的机器。最好的计算机和应用程序设计源自优雅的概念。理论讲座可以增强你的审美意识，帮助你构建更漂亮的系统。

Theory is also important because it shows a new, simpler, and more elegant side to computers, which are usually considered very complex machines. The best computer and application designs emerge from elegant concepts. Theoretical instruction can enhance the aesthetic sense and help build more beautiful systems.

最后，理论是好的，因为研究它可以开发思想。计算机技术正在迅速变化。详细的技术知识虽然现在有用，但几年后就会过时。因此，值得培养清晰准确地思考和表达自己的能力，以便能够解决问题并知道我们何时无法解决问题。这些技能具有持久的价值。通过学习理论，我们在这些领域进行实践。

Finally, theory is good because studying it develops the mind. Computer technology changes rapidly. Detailed technical knowledge, although useful now, will become obsolete in a few years. It is therefore worth developing the ability to think, to express oneself clearly and precisely, to be able to solve problems and to know when we have not been able to solve a problem. Such skills have lasting value. By studying theory, we practice in these areas.

除了实际用途问题之外，几乎每个在工作中使用计算机的人都对这些非凡的创造、它们的可能性和局限性感到好奇。在过去的三十年里，一个新的数学分支已经发展起来，研究人员在其中寻找某些基本问题的答案。还有一个悬而未决的问题：如果我们取一个很大的数字（比如 500 位数字），我们能否在合理的时间内找到它的约数（除以它而没有余数的数字）？即使使用超级计算机，目前也没有人知道如何在小于宇宙年龄的时间内针对每种情况执行此操作！因式分解问题（即分解问题）与现代加密系统密切相关。找到一种快速分解数字的方法，您就会出名！

Aside from their practical utility, almost everyone who uses computers at work is curious about these amazing creations, their capabilities, and their limitations. Over the past thirty years, a new branch of mathematics has emerged, whose researchers are trying to answer some fundamental questions. Here is one that remains unsolved: If we take a large number—say, 500 digits—can we find its divisors (the numbers by which it divides without remainder) in a reasonable amount of time? Even with a supercomputer, no one currently knows how to do this for every case in less than the age of the universe ! The factoring problem (or the problem of factoring) is closely related to modern encryption systems. Find a fast way to factor a number, and you will become famous!

致老师们

To teachers

本书旨在作为理论计算机科学领域的教科书，供硕士研究生以及学士和工程研究的高年级学生使用。这些问题是基于定理和证明以数学方式提出的。我已经尽力向那些缺乏定理证明经验的学生介绍，但是更高级的学生会发现它更容易。

This book is intended as a textbook on theoretical computer science for graduate students and advanced undergraduate and graduate students. The topics are presented mathematically, based on theorems and proofs. I have made an effort to introduce students who have little experience in proving theorems, but more advanced students will find it easier.

我的主要目标是以清晰有趣的方式呈现材料。正因为如此，我更喜欢直觉和大局，而不是对底层细节的详细分析。

My main goal was to present the material in a clear and interesting way. For this reason, I favored intuition and the big picture over detailed analysis of lower-level details.

例如，虽然我在第Ø章中介绍了归纳证明方法以及其他数学基础，但它在本书的其余部分中并没有发挥重要作用。一般来说，我不会展示关于自动机的各种结构的正确性的典型归纳证明。如果表述清楚，这样的解释就可以为自己辩护，不需要进一步的论证。归纳法可能会掩盖而不是澄清问题，因为它本身是一种相当复杂的技术，对许多人来说是难以理解的。通过归纳法解释明显的事实存在着让学生永远相信数学证明是形式操纵的一种形式的风险，而不是教他们区分令人信服的论点和不令人信服的论点。

For example, although I have presented the method of proof by induction in Chapter Ø, along with other mathematical foundations, it does not play an important role in the rest of the book. In general, I do not present typical inductive proofs of the correctness of various constructions concerning automata. If they are presented clearly, such constructions are self-defeating and do not require further justification. Induction could obscure rather than clarify the problem, since it is in itself a rather sophisticated technique that many people do not understand. Explaining obvious facts by induction runs the risk of perpetuating the idea that mathematical proof is a kind of formal manipulation, instead of teaching them to distinguish convincing arguments from unconvincing ones.

第二个例子出现在第二部分和第三部分中，我用日常语言而不是使用伪代码描述算法。我没有花太多时间对图灵机（或任何其他正式模型）进行编程。今天的学生都有编程背景，丘奇-图灵假设对他们来说是显而易见的。因此，我不会通过一个模型对另一个模型进行长时间的模拟来证明它们的等效性。

The second example occurs in Parts 2 and 3, where I describe algorithms in everyday language rather than using pseudocode. I have not spent much time programming Turing machines (or any other formal model). Today's students have a programming background, and the Church-Turing thesis is obvious to them. So I do not present long simulations of one model by the other to show their equivalence.

尽管我更重视直觉并省略了某些细节，但我以一种可以称为经典的方式呈现材料。大多数理论家会发现材料的选择、术语和呈现顺序与其他广泛使用的教科书中包含的内容一致。当我发现常用术语特别不清楚或令人困惑时，我只在少数地方引入了我自己原创的术语。例如，我引入了映射归约的概念，而不是多对一归约。

Although I give more weight to intuition and omit some details, I present the material in a way that may be called classical. Most theorists will find that the choice of material, the terminology, and the order of presentation are consistent with those in other widely used textbooks. I have introduced my own original terminology only in a few places, where I found the standard concepts to be particularly obscure or confusing. For example, I have introduced the notion of reduction by mapping in place of many-to-one reduction .

通过解决问题进行练习是学习任何数学领域的基础。书中提出的任务分为两类：练习和任务。这些练习的目的是回忆和巩固定义和概念。这些任务需要一定的创造力。标有星号的任务更加困难。我试图让解决练习和任务成为对学生来说有趣的挑战。

Practicing by solving problems is the basis for learning any mathematical discipline. The problems presented in the book are divided into two categories: Exercises and Problems. The exercises are intended to recall and consolidate definitions and concepts. The problems require some creativity. Problems marked with an asterisk are more difficult. I tried to make solving both exercises and problems an interesting challenge for students.

第一版

First edition

《计算理论导论》首次以平装介绍性版本出版。第一版与初步版在几个重要方面有所不同。添加了最后三章：第 8 章关于内存复杂性，第 9 章关于不可计算问题，第 10 章关于复杂性理论的高级主题。第 6 章经过扩展，包含了可计算性理论中的几个广泛主题。其他章节也补充了额外的示例和练习。

Introduction to Computation Theory was first published as a paperback introductory edition. The first edition differs from the introductory edition in several important respects. Three final chapters have been added: Chapter 8 on memory complexity, Chapter 9 on uncomputable problems, and Chapter 10 on advanced topics in complexity theory. Chapter 6 has been expanded to include several extended topics in computability theory. Other chapters have also been supplemented with additional examples and exercises.

我从使用预印本版本的教师和学生那里收到的评论有助于完善第 0-7 章。当然，我也纠正了引起我注意的错误。

The comments I received from instructors and students using the preprint helped me refine Chapters 0–7. Naturally, I also corrected errors that were brought to my attention.

第 6 章和第 10 章回顾了可计算性和复杂性理论中的一些更高级的主题。它们并不像其他章节那样以全面和连贯的方式呈现主题。我添加这些章节是为了让教师选择学生可能感兴趣的可选主题。相同的问题属于不同的领域。其中一些，例如图灵可约性或交替性，是本书中讨论的其他概念的直接扩展。其他的，例如可判定的逻辑理论或密码学，只是对广泛研究领域的简要介绍。

Chapters 6 and 10 review several more advanced topics in computability and complexity theory. They are not intended to be as comprehensive or coherent as the other chapters. I have included these chapters to allow the teacher to select optional topics that may be of interest to students. The topics themselves belong to a variety of areas. Some, such as Turing reducibility or alternations , are direct extensions of other concepts discussed in the book. Others, such as decidable logical theories or cryptography , are only brief introductions to large areas of study.

作者须知

Notes to the author

互联网为作者和读者之间的互动创造了新的机会。我收到了许多包含建议、赞扬和批评的电子邮件，包括预发布中的错误报告。请继续写！我尝试在时间允许的情况下尽快回复每条消息。有关本书的信件应发送至的地址是

The Internet has opened up new possibilities for interaction between authors and readers. I have received many e-mails containing suggestions, compliments, and criticisms, including reports of errors in the initial edition. Please continue to write! I try to answer every message as time permits. The address for correspondence concerning this book is

sipserbook@math.mit.edu。

sipserbook@math.mit.edu.

包含该书勘误的网站也在持续维护中。该网站还可能包含对教师和学生有用的其他材料。如果您能提供有关在那里可以找到什么的信息，我将不胜感激。此页面位于：

There is also an ongoing website containing errata for the book. Other materials useful for teachers and students may also appear on this website. I would be grateful for information on what could be found there. This website is available at:

http://math.mit.edu/~sipser/book.html。

http://math.mit.edu/~sipser/book.html .

致谢

Thanks

如果没有许多朋友、同事和家人的帮助，我不可能写出这本书。

I could not have written this book without the help of many friends, colleagues and my family.

我要感谢我的老师，他们帮助我塑造了我的科学观点和教学方法。其中有五个特别突出。我的导师曼努埃尔·布鲁姆（Manuel Blum）值得特别提及，因为他具有通过清晰的思维、热情和关怀来激励学生的非凡能力。他是我和许多其他人的榜样。我还感谢理查德·卡普 (Richard Karp) 向我介绍了复杂性理论，感谢约翰·艾迪生 (John Addison) 教我逻辑并创造了出色的作业问题，感谢尤里斯·哈特马尼斯 (Juris Hartmanis) 向我介绍了计算理论，感谢我的父亲教我数学基础知识、计算机和传播知识的艺术。

I would like to thank my teachers who helped shape my scientific views and teaching methods. Five of them stand out in particular. My advisor, Manuel Blum, deserves special mention for his exceptional ability to inspire students through the clarity of his ideas, his enthusiasm, and his care. He is a role model for me and many others. I am also grateful to Richard Karp for introducing me to the secrets of complexity theory, John Addison for teaching me logic and preparing great homework assignments, Juris Hartmanis for introducing me to the theory of computation, and my father for teaching me the basics of mathematics, the use of computers, and the art of communicating knowledge.

这本书是根据我过去 15 年在麻省理工学院所做的一次演讲的笔记编写的。它们是由参加我讲座的学生写下来的。我希望他们能原谅我没有提及他们所有人。多年来与我共事的助理 – Avrim Blum、Thang Bui、Benny Chor、Andrew Chou、Stavros Cosmadakis、Aditi Dhagat、Wayne Goddard、Parry Husbands、Dina Kravets、Jakov Kučan、Brian O'Neill、Ioana Popescu 和 Alex Russell – 帮助编辑和开发这些笔记。他们也是一些练习和任务的作者。

This book is based on lecture notes I have given at MIT over the past 15 years. They were written by students who attended my lectures. I hope they will forgive me for not mentioning them all. The assistants who have worked with me over the years—Avrim Blum, Thang Bui, Benny Chor, Andrew Chou, Stavros Cosmadakis, Aditi Dhagat, Wayne Goddard, Parry Husbands, Dina Kravets, Jakov Kučan, Brian O'Neill, Ioana Popescu, and Alex Russell—helped to edit and expand these notes. They also wrote some of the exercises and problems.

大约三年前，汤姆·莱顿说服我写一本关于计算理论的教科书。我已经思考这个问题有一段时间了，但正是汤姆的论点将理论转化为实践。我很感谢他在写书和其他事情上提供的大量建议。

Almost three years ago, Tom Leighton convinced me to write a textbook on the theory of computation. I had been thinking about it for some time, but it was Tom's reasoning that allowed me to turn the theory into practice. I am grateful to him for his advice on writing books and for many other things.

我要感谢 Eric Bach、Peter Beebee、Cris Calude、Marek Chrobak、Anna Chefter、Guang-Ien Cheng、Elias Dahlhaus、Michael Fischer、Steve Fisk、Lance Fortnow、Henry J 在本书写作期间提供的意见、评论和帮助。弗里德曼、杰克·福、西摩·金斯伯格、奥德·戈尔德赖希、布莱恩·格罗斯曼、大卫·哈雷尔、米夏·霍弗里、Dung T. Huynh、尼尔·琼斯、H.查德·莱恩、凯文·林、迈克尔·路易、西尔维奥·米卡利、村田忠雄、克里斯托斯。 Papadimitriou、Vaughan Pratt、Daniel Rosenband、Brian Scassellati、Ashish Sharma、Nir Shavit、Alexander Shen、Ilya Shlyakhter、Matt Stallmann、Perry Susskind、YC Tay、Joseph Traub、Osamu Watanabe、Peter Widmayer、David Williamson、Derick Wood 和 Charles Yang。

I would like to acknowledge the comments, suggestions, and assistance provided during the writing of this book by Eric Bach, Peter Beebee, Cris Calude, Marek Chrobak, Anna Chefter, Guang-Ien Cheng, Elias Dahlhaus, Michael Fischer, Steve Fisk, Lance Fortnow, Henry J. Friedman, Jack Fu, Seymour Ginsburg, Oded Goldreich, Brian Grossman, David Harel, Micha Hofri, Dung T. Huynh, Neil Jones, H. Chad Lane, Kevin Lin, Michael Loui, Silvio Micali, Tadao Murata, Christos Papadimitriou, Vaughan Pratt, Daniel Rosenband, Brian Scassellati, Ashish Sharma, Nir Shavit, Alexander Shen, Ilya Shlyakhter, Matt Stallmann, Perry Susskind, YC Tay, Joseph Traub, Osamu Watanabe, Peter Widmayer, David Williamson, Derick Wood, and Charles Yang.

以下人员提供了有助于改进本书的额外评论和反馈：Isam M. Abdelhameed、Eric Allender、Shay Artzi、Michelle Atherton、Rolfe Blodgett、Al Briggs、Brian E. Brooks、Jonathan Buss、Jin Yi Cai、Steve Chapel、David Chow、Michael Ehrlich、Yaakov Eisenberg、Farzan Fallah、Shaun Flisakowski、Hjalmtyr Hafsteinsson、CR Hale、Maurice Herlihy、Vegard Holmedahl、Sandy Irani、Kevin Jiang、Rhys Price Jones、James M. Jowdy、David M. Martin Jr.、Manrique Mata -Montero、Ryota Matsuura、Thomas Minka、Farooq Mohammed、Tadao Murata、Jason Murray、Hideo Nagahashi、Kazuo Ohta、Constantine Papageorgiou、Joseph Raj、Rick Regan、Rhonda A. Reumann、Michael Rintzler、Arnold L. Rosenberg、Larry Roske、Max Rozenoer、Walter L. Ruzzo、Sanatan Sahgal、Leonard Schulman、Steve Seiden、Joel Seiferas、Ambuj Singh、David J. Stucki、Jayram S. Thathachar、H. Venkateswaran、Tom Whaley、Christopher Van Wyk、Kyle Young 和 Kyoung Hwan Yun。

The following people provided additional comments and feedback that improved this book: Isam M. Abdelhameed, Eric Allender, Shay Artzi, Michelle Atherton, Rolfe Blodgett, Al Briggs, Brian E. Brooks, Jonathan Buss, Jin Yi Cai, Steve Chapel, David Chow, Michael Ehrlich, Yaakov Eisenberg, Farzan Fallah, Shaun Flisakowski, Hjalmtyr Hafsteinsson, C. R. Hale, Maurice Herlihy, Vegard Holmedahl, Sandy Irani, Kevin Jiang, Rhys Price Jones, James M. Jowdy, David M. Martin Jr., Manrique Mata-Montero, Ryota Matsuura, Thomas Minka, Farooq Mohammed, Tadao Murata, Jason Murray, Hideo Nagahashi, Kazuo Ohta, Constantine Papageorgiou, Joseph Raj, Rick Regan, Rhonda A. Reumann, Michael Rintzler, Arnold L. Rosenberg, Larry Roske, Max Rozenoer,Walter L. Ruzzo, Sanatan Sahgal, Leonard Schulman, Steve Seiden, Joel Seiferas, Ambuj Singh, David J. Stucki, Jayram S. Thathachar, H. Venkateswaran, Tom Whaley, Christopher Van Wyk, Kyle Young, and Kyoung Hwan Yun.

罗伯特·斯隆在他的演讲中使用了这本书的草稿，并从这次经历中给了我宝贵的反馈和想法。马克·赫施伯格 (Mark Herschberg)、太田一夫 (Kazuo Ohta) 和拉塔尼亚·斯威尼 (Latanya Sweeney) 阅读了部分手稿并提出了多次修改建议。 Shafi Goldwasser 帮助我准备了第 10 章的材料。

Robert Sloan used a working version of the book in his lecture and gave me invaluable feedback and ideas from that experience. Mark Herschberg, Kazuo Ohta, and Latanya Sweeney read parts of the manuscript and suggested numerous improvements. Shafi Goldwasser helped me with the material for Chapter 10.

我得到了来自 Superscript 的 William Baxter 的技术帮助，他编写了一个 L ^A T _E X 宏包来实现本书的内部布局，还有麻省理工学院数学系的 Larry Nolan 的技术帮助，他确保一切正常。

I received technical assistance from William Baxter of Superscript, who wrote a L ^A T _E X macro package that implemented the book's internal layout, and from Larry Nolan of the MIT mathematics department, who made sure everything worked.

很高兴与 PWS Publishing 团队合作创建最终版本。我只会提及迈克尔·舒格曼、大卫·迪茨、埃莉斯·凯撒、莫妮克·卡莱罗、苏珊·加兰和塔尼娅·布鲁尔，因为他们是我接触最多的人，但我意识到还有很多其他人也参与其中。感谢 Jerry Moore 的文案编辑、Diane Levy 的封面设计以及 Catherine Hawkes 的内部布局设计。

It was a pleasure to work with the team at PWS Publishing in creating the final version. I will mention Michael Sugarman, David Dietz, Elise Kaiser, Monique Calello, Susan Garland, and Tanja Brull, as they were the ones I had most contact with, but I know that many others were involved. Thanks to Jerry Moore for copyediting, Diane Levy for the cover design, and Catherine Hawkes for the internal layout design.

我们感谢美国国家科学基金会 CCR-9503322 拨款的支持。

We thank the National Science Foundation for support under Grant CCR-9503322.

我的父亲肯尼思·西普瑟和妹妹劳拉·西普瑟将书中的图画转换成电子形式。我的另一个姐姐凯伦·费什 (Karen Fisch) 将我们从各种计算机故障中解救出来，而我的母亲贾斯汀·西普瑟 (Justine Sipser) 则提供了慈母般的建议。我感谢他们所有人在困难的情况下提供的支持，包括疯狂的截止日期和抵制软件。

My father, Kenneth Sipser, and sister, Laura Sipser, converted the drawings in the book into electronic form. My other sister, Karen Fisch, rescued us from various computer troubles, and my mother, Justine Sipser, provided motherly advice. I thank them all for their support under difficult circumstances, including crazy deadlines and stubborn software.

最后，我向我的妻子伊娜和女儿雷切尔表示爱意。谢谢你忍受这一切。

Finally, my love goes out to my wife Ina and daughter Rachel. Thank you for bearing with me through it all.

迈克尔·西普瑟

Michael Sipser

马萨诸塞州剑桥，1996 年 10 月

Cambridge, Massachusetts, October 1996

第二版前言

Preface to the Second Edition

从我从这么多读者那里收到的电子邮件来看，第一版的最大缺点是甚至缺乏某些任务的示例解决方案。所以他们已经在这里了。现在，每一章都包含一个名为“选定解决方案”的新部分，该部分提供了该章中代表性练习和任务选择的答案。为了补充减少的自行解决任务选择，我还添加了许多新的练习和任务。讲师还可以订购包含其他解决方案的教师手册。为此，请联系www.course.com上列出的您的区域销售代表。

Judging from the emails I have received from so many readers, the biggest flaw in the first edition was the lack of sample solutions to at least some of the problems. Well, here they are. Each chapter now contains a new section entitled Selected Solutions , which provides answers to a representative selection of exercises and problems in that chapter. To supplement the reduced selection of problems for self-study, I have also added a number of new exercises and problems. Instructors can also order a Teacher's Manual with additional solutions. To do this, contact your regional sales representative listed on www.course.com .

许多读者要求更广泛地报道一些“标准”主题，特别是迈希尔-尼罗德定理和莱斯定理。我部分地满足了这些要求，将这些问题发展为问题并提出解决方案。我没有在正文中包含Myhill-Nerod定理，因为我认为本次讲座应该只提供有限自动机理论的介绍，而不是对其进行深入分析。在我看来，有限自动机在本书中的作用是为学生提供一个简单的形式化计算模型，作为更强大模型的前奏，并为后面的主题提供方便的示例基础。当然，有些人会更喜欢对这个主题进行更详细的处理，而另一些人则希望我完备省略对有限自动机的任何提及（或者至少不要使后续主题依赖于它们）。我没有将赖斯定理纳入本书的正文中，因为尽管它是证明不可判定性的有用工具，但有些学生可能会机械地使用它而没有真正理解问题。通过约简而不是赖斯定理来证明不可判定性，为复杂性理论中稍后出现的约简提供了更好的准备。

Many readers have requested more extensive treatment of some of the "standard" topics, in particular the Myhill-Nerod theorem and the Rice theorem. I have partially complied with these requests by developing these topics as problems with solutions. I have not included the Myhill-Nerod theorem in the main text, since I feel that this lecture should provide only an introduction to the theory of finite automata, not a thorough treatment of it. In my view, the role of finite automata in this book is to provide students with a simple formal model of computation as a prelude to more powerful models, and also to provide a convenient basis for examples in later topics. Of course, some will prefer a more detailed treatment of this topic, while others would like me to omit all mention of finite automata (or at least not make later topics dependent on them). I have not included Rice's theorem in the main text of this book because, although it is a useful tool for proving undecidability, some students might use it mechanically, without really understanding the nature of the problem. Proving undecidability by reduction rather than by Rice's theorem provides better preparation for the reductions that appear later in complexity theory.

我非常感谢我的助手帮助我开发了一些新任务和解决方案。他们是 Ilya Baran、Sergi Elizalde、Rui Fan、Jonathan Feldman、Venkatesan Guruswami、Prahladh Harsha、Christos Kapoutsis、Julia Khodor、Adam Klivans、Kevin Matulef、Ioana Popescu、April Rasala、Sofya Raskhodnikova 和 Iuliu Vasilescu。 Ching W. Law、Edmond Kayi Lee 和 Zulfikar Ramzan 也为解决方案的开发做出了贡献。感谢 Victor Shoup 找到了一种简单的方法来填补第一个版本中存在的概率素性算法分析中的空白。

I am indebted to my assistants who helped me develop some of the new problems and solutions. They are Ilya Baran, Sergi Elizalde, Rui Fan, Jonathan Feldman, Venkatesan Guruswami, Prahladh Harsha, Christos Kapoutsis, Julia Khodor, Adam Klivans, Kevin Matulef, Ioana Popescu, April Rasala, Sofya Raskhodnikova, and Iuliu Vasilescu. Ching W. Law, Edmond Kayi Lee, and Zulfikar Ramzan also contributed to the solutions. I thank Victor Shoup for finding a simple way to patch a gap in the analysis of the probabilistic primality algorithm that existed in the first edition.

感谢 Course Technology 的人们激励我和这个项目的其他参与者，特别是 Alyssia Pratt Aimee Poirier。还要非常感谢 Gerald Eisman、Weizhen Mao、Rupak Majumdar、Chris Umans 和 Christopher Wilson 的审阅。我还要感谢 Jerry Moore 在编辑文本方面所做的出色工作，以及 ByteGraphics 的 Laura Segel (lauras@bytegraphics.com) 提供的精美插图。

Thanks to the people at Course Technology for motivating me and the other participants in this project, especially Alyssia Pratt, Aimee Poirier. Many thanks to Gerald Eisman, Weizhen Mao, Rupak Majumdar, Chris Umans, and Christopher Wilson for their reviews. I am also indebted to Jerry Moore for his excellent work in copyediting, and to Laura Segel of ByteGraphics (lauras@bytegraphics.com) for her beautiful work with the figures.

我收到的电子邮件数量超出了我的预期。收到来自这么多地方的消息是一次很棒的经历，我已经尽力回复每个人 - 请原谅我错过的那些。我在这里提到那些对本次版本产生特别影响的人，但我要感谢大家的来信：

The number of emails I received exceeded my expectations. It was a wonderful experience to receive messages from so many places, and I tried to respond to everyone—I apologize to those I have missed. I include here those whose suggestions were particularly influential in this edition, but I thank all for their correspondence:

卢卡·阿塞托、阿拉什·阿夫坎普尔、罗斯托姆·阿加尼安、埃里克·阿伦德、卡伦·巴克什、布拉德·巴林格、雷·巴特库斯、路易斯·巴顿、阿诺德·贝克曼、米希尔·贝拉尔、凯文·特伦特·伯杰森、马修·伯曼、拉杰什·巴特、索梅纳特·比斯瓦斯、莱诺尔·布鲁姆、毛罗·A·博纳蒂保罗·邦丁,尼古拉斯·博恩,伊恩·布拉特,吉恩·布劳德,道格·伯克,萨姆·巴斯,弗拉基米尔·比奇科夫斯基,布鲁斯·卡尼尔,索玛·乔胡里,陈荣杰,萨米尔·乔普拉,本尼·乔,约翰·克劳森,艾莉森·科茨,安妮·康登,杰弗里·康斯丁、约翰·J·克拉谢尔、克劳德·克雷波、肖恩·卡茨、苏谢尔·M·达斯瓦尼、杰夫·戴维斯、斯科特·德克斯特、彼得·德雷克、杰夫·埃德蒙兹、雅科夫·艾森伯格、库尔特塞贝·埃罗格鲁、乔治·埃斯尔、亚历山大·T·费德、法尔赞·法拉、菲斯·费奇、约瑟夫E. Fitzgerald、佩里·菲扎诺、大卫·福特、Jeannie Fromer、Kevin Fu、Atsushi Fujioka、Michel Galley、K. Ganesan、Simson Garfinkel、Travis Gebhardt、Peymann Gohari、Ganesh Gopalakrishnan、Steven Greenberg、Larry Griffith、Jerry Grossman、Rudolf de Haan , 迈克尔·哈尔珀, 尼克·哈维, 麦克·亨德里克斯, 劳里·希亚库莫托, 史蒂夫·霍克马, 迈克尔·赫勒, 沙哈达特·侯赛因, 戴夫·伊塞克, 盖斯·伊萨, Raj D. Iyer, 克里斯蒂安·雅各比, 托马斯·詹森, 迈克·D·琼斯, 马克斯·卡诺维奇, 亚伦·考夫曼, Roger Khazan、Sarfraz Khurshid、Kevin Killourhy、Seungjoo Kim、Victor Kuncak、Kanata Kuroda、Thomas Lasko、Suk Y. Lee、Edward D. Legenski、Li-Wei Lehman、Kong Lei、Zsolt Lengvarszky、Jeffrey Levetin、Baekjun Lim、Karen Livescu , 史蒂芬·路易, Tzer Hung Low, 沃尔夫冈·马斯, 阿拉什·马达尼, 迈克尔·马纳帕特, 沃伊切赫·马尔丘卡, 小大卫·M·马丁, 安德斯·马丁森, 莱尔·麦克乔奇, 阿尔贝托·梅迪纳, 库尔特·梅尔霍恩, 尼哈尔·梅塔, 艾伯特·R·迈耶, 托马斯·敏卡, Mariya Minkova、Daichi Mizuguchi、G. Allen Morris III、Damon Mosk-Aoyama、牟晓龙、Paul Muir、German Muller、Donald Nelson、Gabriel Nivasch、Mary Obelnicki、Kazuo Ohta、Thomas M. Oleson, Jr.、Curtis Oliver、Owen奥齐尔, 雷内·佩拉尔塔, 亚历山大·佩里斯, 霍尔格·彼得森, 德特勒夫·普朗普, 罗伯特·普林斯, 大卫·普里查德, 比娜·里德, 尼古拉斯·莱利, 罗纳德·里维斯特, 罗伯特·罗宾逊, 克里斯蒂·洛克威尔, 菲尔·罗格威, 马克斯·罗泽诺尔, 约翰·拉普夫, 特奥多·罗斯, 拉里·鲁佐,布莱恩·桑德斯、Cem Say、Kim Schioett、乔尔·塞费拉斯、若昂·卡洛斯·塞图巴尔、杰夫·李·塞永、马克·斯坎德拉、鲍勃·斯隆、杰夫·史密斯、马克·L·史密斯、斯蒂芬·史密斯、亚历克斯·C·斯诺伦、盖伊·圣丹尼斯、拉里·斯托克梅尔、 Radu Stoleru、David Stucki、Hisham M. Sueyllam、Kenneth Tam、Elizabeth Thompson、Michel Toulouse、Eric Tria、Chittaranjan Tripathy、Dan Trubow、Hiroki Ueda、Giora Unger、Kurt L. Van Etten、Jesir Vargas、Bienvenido Velez-Rivera、Kobus Vos、Alex Vrenios、Sven Waibel、Marc Waldman、Tom Whaley、Anthony Widjaja、Sean Williams、Joseph N. Wilson、Chris Van Wyk、邢光明、Vee Voon Yee、程永熙、Neal Young、Timothy Yuen、Kyle Yung、张京华，莉拉·佐莱。

Luca Aceto, Arash Afkanpour, Rostom Aghanian, Eric Allender, Karun Bakshi, Brad Ballinger, Ray Bartkus, Louis Barton, Arnold Beckmann, Mihir Bellare, Kevin Trent Bergeson, Matthew Berman, Rajesh Bhatt, Somenath Biswas, Lenore Blum, Mauro A. Bonatti , Paul Bondin, Nicholas Bone, Ian Bratt, Gene Browder, Doug Burke, Sam Buss, Vladimir Bychkovsky, Bruce Carneal, Soma Chaudhuri, Rong-Jaye Chen, Samir Chopra, Benny Chor, John Clausen, Allison Coates, Anne Condon, Jeffrey Considine , John J. Crashell, Claude Crepeau, Shaun Cutts, Susheel M. Daswani, Geoff Davis, Scott Dexter, Peter Drake, Jeff Edmonds, Yaakov Eisenberg, Kurtcebe Eroglu, Georg Essl, Alexander T. Fader, Farzan Fallah, Faith Fich, Joseph E. Fitzgerald, Perry Fizzano, David Ford, Jeannie Fromer, Kevin Fu, Atsushi Fujioka, Michel Galley, K. Ganesan, Simson Garfinkel, Travis Gebhardt, Peymann Gohari, Ganesh Gopalakrishnan, Steven Greenberg, Larry Griffith, Jerry Grossman, Rudolf de Haan, Michael Halper, Nick Harvey, Mack Hendricks, Laurie Hiyakumoto, Steve Hockema, Michael Hoehle, Shahadat Hossain, Dave Isecke, Ghaith Issa, Raj D. Iyer, Christian Jacobi, Thomas Janzen, Mike D . Jones, Max Kanovitch, Aaron Kaufman, Roger Khazan, Sarfraz Khurshid, Kevin Killourhy, Seungjoo Kim, Victor Kuncak, Kanata Kuroda, Thomas Lasko, Suk Y. Lee, Edward D. Legenski, Li-Wei Lehman, Kong Lei, Zsolt Lengvarszky, Jeffrey Levetin, Baekjun Lim, Karen Livescu, Stephen Louie, Tzer Hung Low, Wolfgang Maass, Arash Madani, Michael Manapat, Wojciech Marchewka, David M. Martin Jr., Anders Martinson, Lyle McGeoch, Alberto Medina, Kurt Mehlhorn, Nihar Mehta, Albert R. Meyer, Thomas Minka, Mariya Minkova, Daichi Mizuguchi, G. Allen Morris III, Damon Mosk-Aoyama, Xiaolong Mou, Paul Muir, German Muller, Donald Nelson, Gabriel Nivasch, Mary Obelnicki, Kazuo Ohta , Thomas M. Oleson, Jr., Curtis Oliver, Owen Ozier, Rene Peralta, Alexander Perlis, Holger Petersen, Detlef Plump, Robert Prince, David Pritchard, Bina Reed, Nicholas Riley, Ronald Rivest, Robert Robinson, Christi Rockwell, Phil Rogaway, Max Rozenoer, John Rupf, Teodor Rus, Larry Ruzzo, Brian Sanders, Cem Say, Kim Schioett, Joel Seiferas, Joao Carlos Setubal, Geoff Lee Seyon, Mark Skandera, Bob Sloan, Geoff Smith, Marc L. Smith, Stephen Smith , Alex C. Snoeren, Guy St-Denis, Larry Stockmeyer, Radu Stoleru, David Stucki, Hisham M. Sueyllam, Kenneth Tam, Elizabeth Thompson, Michel Toulouse, Eric Tria, Chittaranjan Tripathy, Dan Trubow, Hiroki Ueda, Giora Unger, Kurt L. Van Etten, Jesir Vargas, Bienvenido Velez-Rivera, Kobus Vos, Alex Vrenios, Sven Waibel, Marc Waldman, Tom Whaley, Anthony Widjaja, Sean Williams, Joseph N. Wilson, Chris Van Wyk, Guangming Xing, Vee Voon Yee, Cheng Yongxi, Neal Young, Timothy Yuen, Kyle Yung, Jinghua Zhang, Lilla Zollei.

还要感谢那些指出第一版中错误的人：Suzanne Balik、Matthew Kane、Kurt L. Van Etten、Nancy Lynch、Gregory Roberts 和 Cem Say。

Thanks are also due to those who pointed out errors in the first edition: Suzanne Balik, Matthew Kane, Kurt L. Van Etten, Nancy Lynch, Gregory Roberts, and Cem Say.

最重要的是，我要感谢我的家人——艾娜（Ina）、雷切尔（Rachel）和亚伦（Aaron）——当我无休止地坐在电脑屏幕前时，他们的耐心、理解和关爱。

Most of all, I would like to thank my family – Ina, Rachel and Aaron – for their patience, understanding and affection as I spent endless hours in front of my computer screen.

迈克尔·西普瑟

Michael Sipser

马萨诸塞州剑桥，2004 年 12 月

Cambridge, Massachusetts, December 2004

第三版前言

Preface to the Third Edition

第三版包括关于确定性上下文无关语言的全新部分。我选择这个主题有几个原因。首先，它填补了先前自动机理论和形式语言方法中的明显空白。旧版本介绍了有限状态自动机和图灵机的确定性和非确定性变体的概念，但我只讨论了堆叠自动机的非确定性版本。添加对确定性堆叠自动机的讨论是拼图中缺失的部分。

The third edition includes an entirely new section on deterministic context-free languages. I chose this topic for several reasons. First, it fills an obvious gap in previous approaches to automata theory and formal languages. Older editions introduced the concepts of finite automata and Turing machines in deterministic and nondeterministic varieties, but in those editions I discussed only the nondeterministic versions of pushdown automata. The addition of a discussion of deterministic pushdown automata is the missing piece of the puzzle.

其次，确定性上下文无关语言的理论是LR( k )语法的基础，它是自动机理论在编程语言和编译器设计中的一个重要且不平凡的应用。该应用程序结合了许多关键概念，包括确定性和非确定性有限自动机的等效性以及上下文无关语法到堆叠自动机的转换，反之亦然，从而产生了高效且美观的解析方法。所以在这里我们实现了理论与实践的真正合作。

Second, the theory of deterministic context-free languages ​​is the basis of LR( k ) grammars, an important and nontrivial application of automata theory to the design of programming languages ​​and compilers. This application combines many key ideas, including the equivalence of deterministic and nondeterministic finite automata and the transformation of context-free grammars to pushdown automata and back, leading to an efficient and beautiful method of parsing. So we have a real interaction of theory and practice.

最后，鉴于该主题作为自动机理论的自然应用的重要性，现有的计算理论教科书似乎没有充分涵盖该主题。我几年前研究过LR( k ) 语法，但我并没有完备理解它们是如何工作的以及它们如何完美地融入确定性上下文无关语言的理论。在撰写本节时，我的目的是为理论家和实践者提供对该领域直观而严谨的介绍，从而有助于更广泛地理解这些问题。不过，我必须补充一点：本章节中的一些内容要求很高，所以第一堂基础理论课的老师可以选择它作为补充阅读。后面的章节不依赖或参考本子章节的内容。

Finally, this topic seems to be insufficiently covered in existing textbooks on the theory of computation, given its importance as a natural application of automata theory. I studied LR( k ) grammars years ago, but I did not fully understand how they work and did not see how beautifully they fit into the theory of deterministic context-free languages. In writing this section, I have aimed to provide an intuitive but rigorous introduction to the field for both theorists and practitioners, and thus to contribute to a wider appreciation of the subject. I must add a warning, however: some of the material in this section is quite demanding, so a teacher giving a first, basic lecture on theory may choose it as additional reading. Later chapters do not depend on, and do not refer to, the content of this section.

许多人直接或间接地为这个版本的创作做出了贡献。我感谢审稿人 Christos Kapoutsis 和 Cem Say，他们阅读了新部分的草稿并分享了宝贵的意见。来自 Cengage Learning 的许多人都参与了制作，尤其是艾莉莎·普拉特 (Alyssa Pratt) 和詹妮弗·费尔特里-乔治 (Jennifer Feltri-George)。 Suzanne Huizenga 编辑了文本，ByteGraphics 的 Laura Segel 创建了新插图并修改了一些旧绘图。

Many people contributed directly or indirectly to this edition. I am grateful to reviewers Christos Kapoutsis and Cem Say, who read the draft of the new part and provided valuable feedback. Many people at Cengage Learning contributed to the production, especially Alyssa Pratt and Jennifer Feltri-George. Suzanne Huizenga edited the text, and Laura Segel of ByteGraphics provided new illustrations and modified some of the older figures.

我要感谢我的麻省理工学院助理 Victor Chen、Andy Drucker、Michael Forbes、Elena Grigorescu、Brendan Juba、Christos Kapoutsis、Jon Kelner、Swastik Kopparty、Kevin Matulef、Amanda Redlich、Zack Remscrim、Ben Rossman、Shubhangi Saraf 和 Oren Weimann 。他们每个人都通过讨论新任务及其解决方案来帮助我，并深入了解我们的学生对讲座内容的理解程度。与这些才华横溢、充满热情的年轻人一起工作真的很高兴。

I would like to thank my assistants at MIT, Victor Chen, Andy Drucker, Michael Forbes, Elena Grigorescu, Brendan Juba, Christos Kapoutsis, Jon Kelner, Swastik Kopparty, Kevin Matulef, Amanda Redlich, Zack Remscrim, Ben Rossman, Shubhangi Saraf, and Oren Weimann. Each of them helped me by discussing new problems and their solutions and by giving me insight into how well our students understood the content of the lecture. It is a real pleasure to work with such talented and enthusiastic young people.

最大的奖励是收到来自世界各地的带有注释和评论的电子邮件。感谢您提出的所有建议、问题和想法。以下是影响本版评论的通讯员名单：

It is a great reward to receive emails with comments and feedback from all over the world. Thank you for all your suggestions, questions and ideas. Here is a list of those correspondents whose comments have influenced this edition:

吉赫德·阿菲菲,史蒂夫·奥尔德里奇,埃里克·巴克,苏珊娜·巴利克,维克多·班杜尔,保罗·比姆,埃拉扎尔·伯恩鲍姆,古塔姆·比斯瓦斯,罗布·比特纳,玛丽娜·布兰顿,罗德尼·布利斯,普罗米塔·查克拉博蒂,刘易斯·科利尔,乔纳森·德伯,西蒙·德克斯特,马特·迪普豪斯,彼得·迪林杰, Peter Drake, 杜志典, Peter Fejer, 玛格丽特·弗莱克, Atsushi Fujioka, Valerio Genovese, Evangelos Georgiadis, Joshua Grochow, Jerry Grossman, Andreas Guelzow, Hjalmtyr Hafsteinsson, Arthur Hall III, Cihat Imamoglu, Chinawat Isradisaikul, Kayla Jacobs, Flemming Jensen,芭芭拉·凯撒、马修·凯恩、克里斯托斯·卡普西斯、阿里·杜洛夫·汗、埃德温·邱诗伦、金容旭、阿卡什·库马尔、埃利亚撒·莱尔、兹索尔特·伦瓦尔斯基、李正中、梁向东、弗拉基米尔·利夫希茨、瑞恩·洛蒂、乔纳森·洛、南希·林奇、亚历克西斯·马西尔, 凯文·马图莱夫, 纳尔逊·马克斯, 汉斯-鲁道夫·梅斯, 姆拉登·米克·扎,萨拉·迈纳·莫尔, Rajagopal Nagarajan, 马文·中山, 乔纳斯·尼拉普, 格雷戈里·罗伯茨, 瑞安·罗梅罗, 桑托什·萨马蒂亚姆, Cem Say, 乔尔·塞弗拉斯, 约翰·西格,马克·史密斯、约翰·斯坦伯格、努里·塔斯德米尔、塔米尔·塔萨、马克·特斯塔、杰西·张、约翰·特拉梅尔、Hiroki Ueda、Jeroen Vaelen、Kurt L. Van Etten、Guillermo Vazquez、Phanisekhar Botlaguduru Venkata、Benjamin Bing-Yi Wang、Lutz Warnke 、大卫·沃伦、托马斯·沃森、约瑟夫·威尔逊、大卫·维滕伯格、布莱恩·旺乔瓦特、基山·耶鲁班迪、戴毅。

Djihed Afifi, Steve Aldrich, Eirik Bakke, Suzanne Balik, Victor Bandur, Paul Beame, Elazar Birnbaum, Goutam Biswas, Rob Bittner, Marina Blanton, Rodney Bliss, Promita Chakraborty, Lewis Collier, Jonathan Deber, Simon Dexter, Matt Diephouse, Peter Dillinger , Peter Drake, Zhidian Du, Peter Fejer, Margaret Fleck, Atsushi Fujioka, Valerio Genovese, Evangelos Georgiadis, Joshua Grochow, Jerry Grossman, Andreas Guelzow, Hjalmtyr Hafsteinsson, Arthur Hall III, Cihat Imamoglu, Chinawat Isradisaikul, Kayla Jacobs, Flemming Jensen, Barbara Kaiser, Matthew Kane, Christos Kapoutsis, Ali Durlov Khan, Edwin Sze Lun Khoo, Yongwook Kim, Akash Kumar, Eleazar Leal, Zsolt Lengvarszky, Cheng-Chung Li, Xiangdong Liang, Vladimir Lifschitz, Ryan Lortie, Jonathan Low, Nancy Lynch, Alexis Maciel, Kevin Matulef, Nelson Max, Hans-Rudolf Metz, Mladen Mik Ŝ a, Sara Miner More, Rajagopal Nagarajan, Marvin Nakayama, Jonas Nyrup, Gregory Roberts, Ryan Romero, Santhosh Samarthyam, Cem Say, Joel Seiferas, John Sieg, Marc Smith, John Steinberger, Nuri Tas¸demir, Tamir Tassa, Mark Testa, Jesse Tjang, John Trammell, Hiroki Ueda, Jeroen Vaelen, Kurt L. Van Etten, Guillermo Vazquez, Phanisekhar Botlaguduru Venkata, Benjamin Bing-Yi Wang, Lutz Warnke, David Warren, ThomasWatson, Joseph Wilson, David Wittenberg, Brian Wongchaowart, Kishan Yerubandi, Dai Yi.

最重要的是，我要感谢我的家人——我的妻子伊娜（Ina）和我们的孩子雷切尔（Rachel）和亚伦（Aaron）。时间是有限的，而且过得很快。你的爱就是一切。

Above all, I thank my family – my wife Ina and our children Rachel and Aaron. Time is finite and slipping away. Your love is everything.

迈克尔·西普瑟

Michael Sipser

马萨诸塞州剑桥，2012 年 4 月

Cambridge, Massachusetts, April 2012

0

。简介

0

. Introduction

我们将从本书中介绍的计算理论领域开始。然后，读者将有机会学习和/或回忆他或她稍后需要的某些数学概念。

We will begin by reviewing the areas of computational theory that will be covered in this book. Then, the reader will have the opportunity to learn and/or refresh their memory on some of the mathematical concepts that they will need later.

0.1

自动机、可计算性和复杂性

0.1

Automata, Computability and Complexity

本书重点关注传统上与计算理论相关的三个领域：自动机、可计算性和复杂性。这些领域通过以下问题联系起来：

This book focuses on three areas traditionally associated with the theory of computation: automata, computability, and complexity. These areas are united by the question:

计算机的基本功能和局限性是什么？

What are the fundamental capabilities and limitations of computers?

这个问题可以追溯到 20 世纪 30 年代，当时数理逻辑学家首次开始考虑计算的本质。从那时起，技术进步极大地提高了我们的计算能力，这意味着这个问题不再是理论领域，而是成为实际应用领域的一部分。

The question goes back to the 1930s, when mathematical logicians first began to ponder the nature of computation. Technological advances since then have greatly expanded our computational abilities and moved the question from the realm of theory to the world of practical applications.

在自动机、可计算性和复杂性这三个领域中，这个问题有不同的解释，并且答案根据这种解释而变化。在这一介绍性章节之后，我们将在本书的不同部分中探讨每个领域。这里我们将以相反的顺序展示这些部分，因为从最后开始，读者将能够更好地理解开头的情况。

In each of these three areas—automata, computability, and complexity—the question is interpreted differently, and the answers change according to that interpretation. After this introductory chapter, we will examine each of these areas in separate parts of the book. Here we will present the parts in reverse order, because by starting at the end the reader will be better able to understand the conditions at the beginning.

复杂性理论

Complexity theory

计算问题有很多种；有些很容易，有些则很难。例如，排序问题就是简单的问题之一。假设我们需要按升序排列数字列表。即使是一台小型计算机也可以很快地对一百万个数字进行排序。让我们将其与日程问题进行比较。假设我们需要找到整个大学的讲座时间表，满足一些合理的约束，例如不能同时在同一个房间进行两场讲座。规划问题似乎比排序问题困难得多。如果我们有一千个小组，即使使用超级计算机，也可能需要很长时间才能找到最佳时间表。

Computational problems come in all sorts; some are easy, others are really hard. For example, the sorting problem is one of the easy ones. Suppose we need to sort a list of numbers into ascending order. Even a small computer can sort a million numbers fairly quickly. Contrast this with the scheduling problem. Suppose we need to find a schedule of lectures for an entire university, while satisfying some reasonable constraints, such as that no two lectures can be held in the same room at the same time. The scheduling problem seems much harder than the sorting problem. If we have a thousand groups, finding the best schedule could take ages, even with a supercomputer.

是什么让一些问题计算起来困难而另一些问题却很容易？

What makes some problems computationally difficult and others easy?

这是复杂性理论的一个核心问题。重要的是，尽管我们进行了 40 多年的深入搜寻，但我们仍然不知道答案。稍后我们将更详细地探讨这个有趣的问题及其一些变体。

This is a central question in complexity theory. Importantly, we do not know the answer to it, despite intensive research over the past 40 years. We will explore this fascinating question and some of its variants in more detail later.

迄今为止，复杂性理论的一项重要进展是，研究人员发现了一种根据计算难度对问题进行分类的优雅方案。它类似于元素周期表，用于根据其化学性质对它们进行分类。使用这个框架，我们可以提供一种方法来查找某些问题在计算上困难的迹象，即使我们无法证明它们确实如此。

In one of the important advances in complexity theory to date, researchers have discovered an elegant scheme for classifying problems according to their computational difficulty. It is analogous to the periodic table of elements used to classify them according to their chemical properties. Using this scheme, we can provide a method for finding indications that certain problems are computationally difficult, even if we cannot prove that they are.

当面对计算上看似困难的问题时，我们有多种选择。首先，通过了解问题的哪个方面导致了这种困难，我们也许能够改变它，使问题变得更容易解决。其次，结果可能是一个不完美的解决方案就足够了。在某些情况下，找到一个近似最佳值的方法是很容易的。第三，有些问题只在最坏的情况下才困难，但在大多数其他情况下都很容易。根据应用程序的不同，我们可能会对偶尔非常慢但通常运行速度很快的过程感到满意。最后，可以考虑使用替代类型的处理（例如随机计算）来加速特定任务。

When faced with a problem that seems computationally difficult, we have several options. First, by understanding what aspect of the problem is causing the difficulty, we may be able to change it so that the problem becomes easier to solve. Second, it may turn out that an imperfect solution is sufficient. In certain situations, finding one that is only an approximation of the best is easy. Third, some problems are difficult only in the worst case, but are easy in most other situations. Depending on the application, we may be satisfied with a procedure that is occasionally very slow but usually runs quickly. Finally, we can consider alternative types of processing, such as randomized computation, which can speed up certain tasks.

复杂性理论直接影响的应用领域之一是古老的密码学领域。在大多数问题中，计算上简单的问题比困难的问题更受欢迎，因为简单的问题解决起来更便宜。密码学的不同寻常之处在于它要求计算问题变得困难而不是简单。在不知道秘密密钥或密码的情况下，秘密密码应该很难破解。复杂性理论引导密码学家解决计算难题，并在此基础上设计新密码。

One area of ​​application that complexity theory has had a direct impact on is the ancient field of cryptography. In most problems, easy problems are preferred over hard ones because it is cheaper to solve easy problems. Cryptography is unusual in that it requires that computational problems be hard, not easy. Secret ciphers should be difficult to break without knowledge of the secret key or password. Complexity theory has directed cryptographers toward hard problems, from which new ciphers have been designed.

可计算性理论

Computability theory

二十世纪上半叶，库尔特·哥德尔、艾伦·图灵和阿隆佐·丘奇等数学家发现，一些基本问题无法使用计算机解决。这种现象的一个例子是确定某个数学陈述是真是假的问题。这项任务是数学家的日常工作。使用计算机解决这个问题似乎是很自然的，因为它属于数学领域。然而，没有计算机算法可以完成这项任务。

In the first half of the twentieth century, mathematicians such as Kurt Gödel, Alan Turing, and Alonzo Church discovered that certain fundamental problems could not be solved by computers. An example of this phenomenon is the problem of determining whether a mathematical statement is true or false. This is a daily task for mathematicians. It would seem that solving it by computers would be natural, since it lies within the realm of mathematics. However, no computer algorithm can accomplish this task.

这一意义深远的成果的一个后果是关于计算机理论模型的思想的发展，这些思想最终将有助于构建真正的计算机。

One consequence of this profound result was the development of concepts for theoretical models of computers that should ultimately help in the construction of real computers.

可计算性和复杂性理论密切相关。在复杂性理论中，目标是将问题分为简单或困难；同时，在可计算性理论中，问题被分为可解决的问题和不可解决的问题。可计算性理论引入了复杂性理论中使用的许多概念。

Computability and complexity theories are closely related. In complexity theory, the goal is to classify problems as easy or hard; meanwhile, in computability theory, problems are classifiable as those that are solvable or those that are not. Computability theory introduces many concepts used in complexity theory.

自动机理论

Theory of automata

自动机理论涉及计算数学模型的定义和属性。这些模型在计算机科学的许多应用领域发挥着作用。其中一种模型称为有限状态机，用于文本处理、编译器和硬件的设计。另一种模型称为上下文无关语法，用于编程语言和人工智能设计。

Automata theory is concerned with the definitions and properties of mathematical models of computation. These models play a role in many applied areas of computer science. One of these models, called a finite-state automaton , is used in the design of text processing, compilers, and hardware. Another model, called a context-free grammar , is used in programming languages ​​and the design of artificial intelligence.

自动机理论是开始研究计算理论的好地方。可计算性和复杂性理论需要对计算机进行精确定义。自动机理论允许您练习计算的正式定义，因为它引入了与计算机科学其他非理论领域相关的想法。

Automata theory is an excellent place to start studying the theory of computation. Theories of computability and complexity require precise definitions of what a computer is. Automata theory allows for practice in formal definitions of computation, as it introduces ideas relevant to other, non-theoretical areas of computer science.

0.2

数学概念和术语

0.2

Mathematical Concepts and Terminology

与任何数学教科书一样，我们将首先讨论我们将使用的基本数学对象、工具和符号。

As with any math textbook, we will begin by discussing the basic mathematical objects, tools, and notations we will be using.

收成

Collections

集合是被视为一个单元的一组对象。集合可以包含任何类型的对象，包括数字、符号，甚至其他集合。集合中的对象称为元素。集合可以用多种方式进行形式化描述。

A set is a group of objects treated as a unit. Sets can contain any type of object, including numbers, symbols, and even other sets. The objects in a set are called elements . Sets can be formally described in a variety of ways.

第一个是在花括号中枚举集合的元素。于是，一个集合

The first one is to enumerate the elements of the set in curly brackets. Thus, the set

S = {7, 21, 57}

S = {7, 21, 57}

包含元素 7、21 和 57。符号ε和∉分别表示属于和不属于该集合。我们可以写成 7 ∈ {7, 21, 57} 和 8 ∉ {7, 21, 57}。对于两个集合A和 B，我们说A是集合B的子集，如果集合A的每个元素也是集合B的元素，我们将其写为A  ⊆  B。如果A是B的子集且不等于B ，则称A是B的真子集，记作A  ⊊ B 。  

contains elements 7, 21, and 57. The symbols ∈ and ∉ denote membership and non-membership, respectively. We can write 7 ∈ {7, 21, 57} and 8 ∉ {7, 21, 57}. For two sets A and  B, we say that A is a subset of B , which we write A  ⊆  B , if every element of A is also an element of B . We say that A is a proper subset of B , which we write A  ⊊  B , if A is a subset of B and not equal to B .

集合中元素的顺序并不重要，元素的重复也不重要。通过写成{57,7,7,7,21}，我们将得到相同的集合S。如果我们想考虑元素出现的次数，我们将这样的组称为多重集，而不是集合。因此，{7} 和 {7, 7} 是不同的多重集，但它们作为集合是相同的。无限集合包含无限多个元素。不可能列出无限集合的所有元素，因此有时我们会使用符号“...”来表示“无限期地继续序列”。因此，我们可以将自然数 集合N写为

The order of the elements in the set is not important, nor is the repetition of the elements. Writing {57, 7, 7, 7, 21} gives us the same set S . If we wanted to take into account the number of occurrences of the elements, we would call such a group a multiset , not a set. Thus, {7} and {7, 7} are different multisets, but they are identical as sets. An infinite set contains infinitely many elements. It is impossible to list all the elements of an infinite set, so sometimes we will use the notation "..." to mean "to continue the sequence into infinity". Thus, we can write the set of natural numbers N as

{1,2,3,…}。

{1, 2, 3, …}.

整数集合Z可以 写为

The set of integers Z can be written as

{…,−2,−1, 0, 1, 2, …}。

{… ,−2,−1, 0, 1, 2, …}.

具有零个元素的集合称为空集，记为Ø。只包含一个元素的集合有时称为单例，包含两个元素的集合有时称为无序对。

A set that has zero elements is called the empty set and is written as Ø . A set containing exactly one element is sometimes called a singleton , and a set with two elements is called an unordered pair .

当我们想要描述一个包含满足特定规则的元素的集合时，我们将其写为{ n ：关于n的规则}。因此，{ n : n = m ² for some m ∈ N } 表示构成自然数平方的数字集合。

When we want to describe a set containing elements satisfying a certain rule, we write it as { n : rule about n }. Thus, { n : n = m ² for some m ∈ N } denotes the set of numbers that are the squares of the natural numbers.

给定两个集合A和 B，它们的和写为A ∪ B ，是我们通过将集合A和 B的所有元素组合成一个而获得的集合。集合 A 和 B的交集（公共部分） ，写作A ∩ B ，是同时属于A和B的元素的集合。集合A的补集，写作A ，是所考虑的域中不属于A 的所有元素的集合。

Given two sets A and  B , their union , written as A ∪ B , is the set obtained by combining all the elements of sets A and  B into one. The intersection ( common part ) of sets A and  B , written as A ∩ B , is the set of those elements that belong to both A and B . The complement of set A , written as A , is the set of all the elements of the domain under consideration that do not belong to A .

正如数学中常见的情况一样，绘图可以帮助阐明某些概念。对于集合，我们使用所谓的维恩图。它们将集合表示为包含在椭圆线内的区域。令集合 START- t为以字母t开头的所有单词的集合。在图中的这个例子中，圆圈代表集合START- t。该集合的几个元素表示为该圆内的点。

As often happens in mathematics, a drawing can help explain some concepts. For sets, we use so-called Venn diagrams . They represent sets as regions enclosed by oval lines. Let the set START- t be the set of all words starting with the letter t . In this example, a circle in the drawing denotes the set START- t . Several elements of this set are represented as points inside the circle.

图 0.1

以字母“ t ”开头的一组单词的维恩图

Figure 0.1

Venn diagram for the set of words starting with the letter " t "

类似地，在下图中，我们可以呈现由以字母“z”结尾的单词组成的 END-z 集合。

Similarly, in the next figure we can present the set END-z consisting of words ending with the letter "z".

图 0.2

以字母“z”结尾的英语单词集的维恩图

Figure 0.2

Venn diagram for the set of English words ending with the letter "z"

为了在同一个维恩图中表示这两个集合，我们需要将它们绘制为使圆圈重叠，表明它们具有一些共同元素，如下图所示。例如，单词“topaz”就属于这两个集合。该图还包含 START- j集的圆圈。它不会与 START -t集的圆圈重叠，因为没有单词可以同时属于这两个集。

To represent both sets in the same Venn diagram, we need to draw them so that the circles overlap, indicating that they have some elements in common, as in the following figure. For example, the word topaz belongs to both sets. This figure also contains a circle for the set START- j . It does not overlap the circle for the set START- t , since no word can belong to both sets.

图 0.3

重叠的圆圈表示共同元素

Figure 0.3

Overlapping circles indicate common elements

接下来的两个维恩图说明了集合A和B 的并集和交集。

The next two Venn diagrams illustrate the union and intersection of sets A and  B .

图 0.4 (a) A ∪ B和 ( b ) A ∩ B

的图表

Figure 0.4

Diagrams for (a) A ∪ B and ( b ) A ∩ B

字符串和元组

Strings and tuples

序列是按特定顺序排列的某些对象的列表。通常通过将字符串的元素括在圆括号中来编写字符串。例如，字符串 7, 21, 57 可以写为

A sequence is a list of certain objects in a specific order. It is customary to write a sequence by placing its elements in parentheses. For example, the sequence 7, 21, 57 can be written as

（7、21、57）。

(7, 21, 57).

在集合中，元素的顺序并不重要，但在序列中，元素的顺序很重要。因此，(7,21,57)与(57,7,21)不是同一个序列。同样，序列中元素的重复很重要，尽管它在集合中无关紧要。这意味着序列 (7, 7, 21, 57) 与之前的两个序列不同，尽管集合 {7, 21, 57} 与集合 {7, 7, 21, 57} 相同。

In a set, the order of elements is irrelevant, but in a sequence it is significant. Thus, (7, 21, 57) is not the same sequence as (57, 7, 21). Similarly, the repetition of elements in a sequence is significant, but it is not significant in a set. This means that the sequence (7, 7, 21, 57) is different from both of the previous sequences, although the set {7, 21, 57} is identical to the set {7, 7, 21, 57}.

与集合一样，序列可以是有限的或无限的。有限序列通常称为元组。具有k 个元素的序列称为k 元组。所以 (7, 21, 57) 是一个三元组。 2 元组通常称为有序对- 类似地，我们使用有序三、四或五的概念。

As with sets, sequences can be finite or infinite. Finite sequences are often called tuples . A sequence with k elements is called a k-tuple . Thus, (7, 21, 57) is a 3-tuple. A 2-tuple is often called an ordered pair – we use the terms 3, 4, or 5 in analogy.

集合和字符串可以显示为其他集合和字符串的成员。例如，A的幂集是A的所有子集的集合。如果A是集合 {0, 1}，则幂集A就是集合 { Ø , {0}, {1}, {0, 1} }。所有元素为 0 和 1 的有序对的集合为 { (0, 0), (0, 1), (1, 0), (1, 1) }。

Sets and sequences can appear as elements of other sets and sequences. For example, the power set of A is the set of all subsets of A . If A is the set {0, 1}, then the power set of A is the set { Ø , {0}, {1}, {0, 1} }. The set of all ordered pairs whose elements are zeros and ones is { (0, 0), (0, 1), (1, 0), (1, 1) }.

如果A和 B是集合，那么它们的笛卡尔积，写作A × B，是所有有序对的集合，其中第一个元素属于集合A，第二个元素属于集合B。

If A and  B are sets, then their Cartesian product , written A × B , is the set of all ordered pairs such that the first element belongs to set A and the second element belongs to set B .

示例0.5

Example 0.5

如果A = {1, 2}且 B = { x , y , z }，则

If A = {1, 2} and B = { x , y , z }, then

A × B = { (1, x ), (1, y ), (1, z ), (2, x ), (2, y ), (2, z ) }。

A × B = { (1, x ), (1, y ), (1, z ), (2, x ), (2, y ), (2, z ) }.

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我们还可以对k个集合A ₁ , A ₂ , … , A _k进行笛卡尔积，写为A ₁ × A ₂ × … × A _k。它是由所有k元组 ( a ₁ , a ₂ , …, a _k )组成的集合，其中a _i ∈ A _i。

We can also perform the Cartesian product of k sets A ₁ , A ₂ , … , A _k , written as A ₁ × A ₂ × … × A _k . This is the set consisting of all k -tuples ( a ₁ , a ₂ , … , a _k ) with a _i ∈ A _i .

示例 0.6

Example 0.6

如果 A 和 B是像示例 0.5 中那样的集合，则

If A and  B are sets such as in Example 0.5, then

A × B × A = { (1, x , 1), (1, x , 2), (1, y , 1), (1, y , 2), (1, z , 1), (1,与, 2),

A × B × A = { (1, x , 1), (1, x , 2), (1, y , 1), (1, y , 2), (1, z , 1), (1, with , 2),

(2, x , 1), (2, x , 2), (2, y , 1), (2, y , 2), (2, z , 1), (2, z , 2)}。

(2, x , 1), (2, x , 2), (2, y , 1), (2, y , 2), (2, z , 1), (2, z , 2)}.

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如果我们创建一个集合与其自身的笛卡尔积，我们可以使用简写符号

If we create the Cartesian product of a set with itself, we can use the shorthand notation

k

k

A × A × … × A = A ^k。

A × A × … × A = A ^k .

示例 0.7

Example 0.7

集合N ²等于N × N。​它由所有有序的自然数对组成。我们也可以将其写为{( i , j ) : i , j ³ 1}。

The set N ² is equal to N × N . It consists of all ordered pairs of natural numbers. We can also write it as {( i , j ) : i , j ³ 1}.

功能和关系

Functions and relationships

函数是数学的基础。函数是定义输入输出关系的对象。它接受输入（参数）并返回结果（值或函数输出）。在每个函数中，相同的参数必须始终返回相同的值。如果f是一个函数，其参数a的输出值为b，我们将其写为

Functions are the foundation of mathematics. A function is an object that defines an input-output relationship. It takes an input (an argument) and returns a result (a value, or the output of the function). In any function, the same argument must always return the same value. If f is a function whose output is b for an argument a , we write this as

f ( a ) = b。

f ( a ) = b .

函数有时也称为映射：如果f ( a )= b，我们说f将a映射到b。

A function is also called a map : if f ( a ) = b , we say that f maps from a to b .

例如，绝对值函数abs将数字x作为输入，如果x是正数则返回x ，如果x是负数则返回−x。因此，abs (2) = abs (−2) = 2。函数的另一个例子可能是加法，我们将其写为add。加法函数的输入是一对有序的数字，输出（值）是这些数字的和。

For example, the absolute value function abs takes a number x as input and returns x if x is positive, and − x if x is negative. Thus, abs (2) = abs (−2) = 2. Another example of a function is addition, which we write as add . The input to the addition function is an ordered pair of numbers, and the output (value) is the sum of those numbers.

函数的可能输入值的集合称为它的域。结果（输出值）属于定义为函数的反域（值集）的集合。 f是一个具有定义域D且值来自集合W 的函数的语句写为

The set of possible input values ​​of a function is called its domain . The results (output values) belong to the set defined as the codomain ( set of values) of the function. The statement that f is a function with domain D and values ​​from the set W , is written as

f : D → W。

f : D → W .

对于函数abs，如果我们将自己限制为整数，则域和子域都是Z，因此我们可以写abs：Z → Z。在整数加法函数的情况下，定义域是整数对Z ×  Z 的集合，反定义域是Z，因此我们将写成add：Z × Z → Z 。请注意，函数不必使用指定值集的所有元素。例如，函数abs永远不会取值 -1，即使 -1 ∈ Z。使用一组值的所有元素的函数称为满射函数。

In the case of the function abs , if we restricted ourselves to integers, both the domain and codomain are Z , so we can write abs : Z → Z . In the case of the integer addition function, the domain is the set of pairs of integers Z  ×  Z , and the codomain is Z , so we can write add : Z × Z → Z . Note that the function does not have to use all the elements of the specified set of values. For example, the function abs never takes the value −1, even though −1 ∈ Z . A function that uses all the elements of the set of values ​​is called an “onto” ( surjective ) function .

函数可以用多种方式表示。一种是计算给定输入值的输出值的过程。另一种方法是使用数组，枚举所有可能的参数并为每个参数提供值。

Functions can be represented in many ways. One is a procedure that calculates an output value for a given input value. Another method is an array that lists all possible arguments and provides values ​​for each.

示例 0.8

Example 0.8

考虑函数f：{0, 1, 2, 3, 4}→{0, 1, 2, 3, 4}。

Consider the function f : {0, 1, 2, 3, 4}→{0, 1, 2, 3, 4}.

n n	f ( n ) f ( n )
0 0	1 1
1 1	2 2
2 2	3 3
3 3	4 4
4 4	0 0

此函数将参数加 1，并返回模 5 的结果。回想一下，模m 的数字是该数字除以m时的余数。例如，钟盘上的分针给出模 60 的值。当使用余数算术时，我们定义Z _m = {0, 1, 2, …, m − 1}。通过这种表示法，前面提到的函数f的形式为f : Z ₅  →  Z ₅。

This function adds 1 to its argument and returns the result modulo 5. Recall that a number modulo m is the remainder when dividing that number by m . For example, the minute hand on a clock face gives values ​​modulo 60. When using remainder arithmetic, we define Z _m = {0, 1, 2, …, m − 1}. With this notation, the previously mentioned function f has the form f : Z ₅  →  Z ₅ .

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示例0.9

Example 0.9

有时，当函数的域是两个集合的笛卡尔积时，会使用二维数组来表示其值。这是另一个函数g : Z ₄  ×  Z ₄  →  Z ₄。行标记为i且列标记为j 的数组单元格包含值g ( i , j )。

Sometimes, when the domain of a function is the Cartesian product of two sets, a two-dimensional array is used to represent its values. Here is another function, g : Z ₄  ×  Z ₄  →  Z ₄ . The array cell in the row labeled i and the column labeled j contains the value g ( i , j ).

G g	0 0	1 1	2 2	3 3
0 0	0 0	1 1	2 2	3 3
1 1	1 1	2 2	3 3	0 0
2 2	2 2	3 3	0 0	1 1
3 3	3 3	0 0	1 1	2 2

g函数是模 4 的加法函数。

The g function is the addition function modulo 4.

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如果对于某些集合A ₁ , … , A _k ，函数的域是A ₁ ×· … ·× A _k ，则f 的输入是k元组 ( a ₁ , a ₂ , … , a _k ) 并且个体a _i称为函数f的参数。具有k 个参数的函数称为k 参数函数，k称为该函数的可论证性（有时称为aryness ）。如果k为 1，则 f有一个参数，称为一元函数。如果k为 2，则 f是双参数（二元）函数。某些众所周知的双参数函数是使用特定的中缀表示法编写的，其中函数符号放置在两个参数之间，而不是通常使用的前缀表示法，其中函数符号位于参数之前。例如，加法函数add通常采用中缀表示法，参数之间加 + 符号，即a + b，而不是前缀表示法add ( a , b )。

If the domain of a function is A ₁ × … × A _k for some sets A ₁ , … , A _k , the input f is a k -tuple ( a ₁ , a ₂ , … , a _k ) and the individual a i _are called the arguments of f . A function with k arguments is called a k-argument function , and k is called the arity of the function. If k is 1, f has a single argument and is called a unary function . If k is 2, f is a binary function . Certain well-known binary functions are written using a special infix notation , in which the function symbol is placed between the two arguments, rather than the more commonly used prefix notation , where the function symbol precedes the arguments. For example, the addition function add is conventionally written in infix notation with a + symbol between its arguments, i.e. a + b , rather than the prefix notation add ( a , b ).

谓词或属性是一个函数，其反域是集合 {TRUE, FALSE}。例如，设Even为属性，如果参数为偶数，则计算结果为 TRUE；如果参数为奇数，则计算结果为 FALSE。因此，即使(4) = TRUE，甚至(5) = FALSE。

A predicate or property is a function whose codomain is the set {TRUE, FALSE}. For example, let even be a property that evaluates to TRUE if the argument is an even number and FALSE if it is an odd number. Thus, even (4) = TRUE and even (5) = FALSE.

定义域为k元组A × ... × A的集合的属性称为关系、k 参数关系或A上的 k 参数关系。一个经常遇到的情况是双参数关系，也称为二元关系。在编写使用二元关系的表达式时，我们通常使用中缀表示法。例如，“小于”是一种传统上使用中缀符号 < 书写的关系。使用符号 = 书写的“平等”是另一种众所周知的关系。如果R是二元关系，则表达式aRb表示aRb = TRUE。类似地，如果R是k参数关系，则表达式R ( a ₁ , …, a _k ) 意味着R ( a ₁ , …, a _k ) = TRUE。

A property whose domain is the set of k -tuples A × … × A is called a relation , a k-argument relation , or a k-argument relation on A . A common case is a binary relation, also called a binary relation. When writing expressions involving binary relations, we customarily use infix notation. For example, “less than” is a relation, which is customarily written using the infix symbol <. “Equality,” written using the symbol =, is another well-known relation. If R is a binary relation, the expression aRb means that aRb = TRUE. Similarly, if R is a k -argument relation , the expression R ( a ₁ , …, a _k ) means that R ( a ₁ , …, a _k ) = TRUE.

示例 0.10

Example 0.10

在石头、剪刀、布的游戏中，两个玩家同时从集合{剪刀、布、石头}中选择一个元素，并用他们的手的适当位置提出他们的选择。如果选择相同，则重复游戏。如果选择不同，则根据胜率，由一名玩家获胜。

In the game of rock, paper, scissors, two players simultaneously select an element from the set {SCISSORS, PAPER, ROCK} and represent the selection by appropriately positioning their hands. If the selections are identical, the game is repeated. If the selections differ, one player wins according to the relationship of captures .

节拍 it beats	剪 SHEARS	纸 PAPER	石头 STONE
剪 SHEARS	错误的 FALSE	真的 TRUE	错误的 FALSE
纸 PAPER	错误的 FALSE	错误的 FALSE	真的 TRUE
石头 STONE	真的 TRUE	错误的 FALSE	错误的 FALSE

从该表中我们可以看出 SCISSORS击败PAPER 为 TRUE，PAPER 击败 SCISSORS 为 FALSE。

From this table we can read that SCISSORS beats PAPER is TRUE, while PAPER beats SCISSORS is FALSE.

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有时将谓词表示为集合而不是函数更方便。谓词P : D →{TRUE, FALSE} 可以写为 ( D , S )，其中S = { a ∈ D : P ( a ) = TRUE} 或简单地写为S （如果D的域从上下文中显而易见）。因此，节拍关系可以写为

It is sometimes more convenient to represent predicates as sets rather than as functions. The predicate P : D →{TRUE, FALSE} can be written as ( D , S ), where S = { a ∈ D : P ( a ) = TRUE} or simply S , if the domain of D is obvious from the context. Thus, the bisecting relation can be written as

{（剪刀，布），（布，石头），（石头，剪刀）}。

{(SCISSORS, PAPER), (PAPER, ROCK), (ROCK, SCISSORS)}.

一种特殊类型的二元关系称为等价关系，它提出了两个对象在某些方面相等的概念。二元关系R是等价关系，如果它满足三个条件：

A special type of binary relation, called an equivalence relation , presents the notion of equality of two objects in some respect. A binary relation R is an equivalence relation if it satisfies three conditions:

1. R是自反的，即对于每个x都有xRx；
2. R is reflexive , that is, for every x, xRx holds ;
3. R是对称的，即对于所有x和y，xRy产生yRx；和
4. R is symmetric , that is, for every x and y , xRy implies yRx ; and
5. R是传递性的，即对于所有x、y和z，如果xRy和yRz，则xRz。
6. R is transitive , that is, for every x , y, and z , if xRy and yRz , then xRz .

示例 0.11

Example 0.11

让我们定义自然数的等价关系，写为≡ ₇。对于i , j  Π N令i == ₇ j，如果i − j是 7 的倍数。这是一个等价关系，因为它满足上述三个条件。首先，它是自反的，因为i − i = 0，它是 7 的倍数。其次，它是对称的，因为如果i − j是 7 的倍数，那么我们可以检查j − i也是 7 的倍数第三，它是传递性的，因为如果i − j是 7 的倍数且j − k是 7 的倍数，则i − k = ( i − j ) + ( j − k ) 是两个倍数之和。 7，因此也是7的倍数。

Let us define an equivalence relation for the natural numbers, written as ≡ ₇ . For i , j  Π N let i ≡ ₇ j , if i − j is a multiple of 7. This is an equivalence relation because it satisfies the three conditions mentioned. First, it is reflexive, since i − i = 0, which is a multiple of 7. Second, it is symmetric, since if i − j is a multiple of 7, then one can check that j − i is also a multiple of 7. Third, it is transitive, since if i − j is a multiple of 7 and j − k is a multiple of 7, then i − k = ( i − j ) + ( j − k ) is the sum of two multiples of 7, and is therefore also a multiple of 7.

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图表

Graphs

无向图或简称图是一组点，其中一些点通过线相互连接。这些点称为顶点或节点，线称为边，如下图所示。

An undirected graph , or simply a graph , is a set of points with lines connecting some of them. These points are called vertices or nodes , and the lines are called edges , as in the figure below.

图 0.12

图形示例

Figure 0.12

Examples of graphs

在特定顶点相交的边的数量称为该顶点的度。在图0.12(a)中，所有顶点的度数为2。在图0.12( b )中，所有顶点的度数为3。任意两个顶点之间最多允许有一条边。在某些情况下，我们可能允许从顶点到自身的边，即环回。

The number of edges meeting at a particular vertex is called the degree of that vertex. In Figure 0.12(a) all vertices have degree 2. In Figure 0.12( b ) all vertices have degree 3. At most one edge is allowed between any two vertices. In some situations we may allow edges leading from a vertex to itself, i.e., loopbacks .

在包含顶点i和j的图G中，对 ( i , j ) 表示连接这些顶点的边。在无向图中， i和j的顺序并不重要，因此 ( i , j ) 和 ( j , i ) 对表示相同的边。无向边有时使用集合表示法表示为无序对，即 { i , j }。如果V是顶点G的集合，E是边的集合，我们可以将其写为G = ( V ,  E )。我们可以使用图表或更正式地指定集合V和E来表示图形。例如，图0.12(a)中图的形式化描述为

In a graph G containing vertices i and j, the pair ( i , j ) represents an edge connecting these vertices. The order of i and j does not matter in an undirected graph, so the pairs ( i , j ) and ( j , i ) represent the same edge. Undirected edges are sometimes represented as unordered pairs using set notation, that is, { i , j }. If V is the set of vertices of G and E is the set of edges, we can write this as G = ( V ,  E ). We can represent the graph by diagramming or more formally by specifying the sets V and E . For example, the formal description of the graph in Figure 0.12(a) is

({1, 2, 3, 4, 5}, {(1, 2), (2, 3), (3, 4), (4, 5), (5, 1)}),

({1, 2, 3, 4, 5}, {(1, 2), (2, 3), (3, 4), (4, 5), (5, 1)}),

图0.12( b )中图的形式化描述为

and the formal description of the graph in Figure 0.12( b ) is

({1, 2, 3, 4}, {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)})。

({1, 2, 3, 4}, {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}).

图表通常用于表示数据。顶点可以是城市，边可以是连接城市的道路，或者顶点可以代表人，边可以代表人与人之间的友谊。有时，为了方便起见，我们可以标记图的顶点和/或边，这称为标记图。图0.13显示了一个图，其顶点是城市，边用这些城市之间最便宜的直飞航班的成本来描述，如果这样的直飞（直飞）是可能的。

Graphs are often used to represent data. The vertices can be cities and the edges can be roads connecting them, or the vertices can represent people and the edges can represent friendships between them. Sometimes, for convenience, we can label the vertices and/or edges of a graph, which is then called a labeled graph . Figure 0.13 shows a graph whose vertices are cities and the edges are labeled with the cheapest direct flight between the cities, if such a direct (non-stop) flight is possible.

图 0.13

不同城市之间最便宜的直飞机票

Figure 0.13

The cheapest non-stop airfares between different cities

如果G 的顶点是H的顶点的子集并且G的边是对应顶点¹的 H的边，我们就说图G是图H的子图。下图显示了图H和子图G。

We say that a graph G is a subgraph of a graph H if the vertices of G are a subset of the vertices of H and the edges of G are the edges of H for the corresponding vertices ¹ . The following figure shows a graph H and a subgraph of G .

图 0.14

图G （显示较暗）是图H的子图

Figure 0.14

The graph G (shown in darker shade) is a subgraph of the graph H

图中的路径是由边连接的顶点序列。直线路径是没有重复顶点的路径。如果对于任何两个顶点都有一条连接它们的路径，则图是连通的。如果一条路径是封闭的，即在同一顶点开始和结束，则该路径称为循环。简单循环是由至少三个不同顶点组成的循环，其中仅重复第一个顶点（作为最后一个顶点）。如果图是连通的且不包含简单循环，则图是树，如图 0.15( c ) 所示。一棵树可能包含一个特殊的顶点，称为根。在树中，除了根之外，度数为1的顶点称为树的叶子。

A path in a graph is a sequence of vertices connected by edges. A simple path ( road ) is a path in which no vertex is repeated. A graph is connected if for any two vertices there is a path connecting them. A path is called a cycle if it is closed, i.e. it starts and ends at the same vertex. A simple cycle is a cycle consisting of at least three different vertices in which only the first vertex (as the last) is repeated. A graph is a tree if it is connected and does not contain simple cycles, as in Figure 0.15 ( c ). A tree may contain a specially distinguished vertex called the root . In a tree, except for the root, the vertices of degree 1 are called leaves of the tree.

图 0.15

(a)图中的路径，( b )图中的循环和( c )树

Figure 0.15

(a) A path in a graph, ( b ) a cycle in a graph, and ( c ) a tree

有向图使用箭头而不是直线表示，如下图所示。来自特定顶点的箭头的数量称为该顶点的输出度，指向给定顶点的箭头的数量称为其输入度。

A directed graph is represented using arrows instead of lines, as shown in the next figure. The number of arrows leaving a given vertex is called the outdegree of that vertex, and the number of arrows pointing to a given vertex is called its indegree .

图 0.16

有向图

Figure 0.16

Directed graph

在有向图中，从顶点i到顶点j 的边表示为有序对 ( i , j )。有向图G的形式表示为 ( V , E )，其中V是顶点集，E - 边集。图 0.16 所示图表的正式表示法是

In a directed graph, an edge from vertex i to vertex j is represented as an ordered pair ( i , j ). The formal notation of a directed graph G is ( V , E ), where V is a set of vertices and E is a set of edges. The formal notation of the graph shown in Figure 0.16 is

({1,2,3,4,5,6}, {(1,2), (1,5), (2,1), (2,4), (5,4), (5,6 ), (6,1), (6,3)})。

({1,2,3,4,5,6}, {(1,2), (1,5), (2,1), (2,4), (5,4), (5,6 ), (6,1), (6,3)}).

从开始到结束所有箭头都指向同一方向的路径称为有向路径。如果对于每两个节点都有一条连接它们的有向路径，则有向图是强连接的。有向图是描述二元关系的便捷方法。如果R是域D × D的二元关系，则它可以表示为标记图G = ( D , E )，其中E = {( x , y ) : xRy }。

A path in which all the arrows point in the same direction, from start to finish, is called a directed path . A directed graph is strongly connected if for every two nodes there is a directed path connecting them. Directed graphs are a convenient way to describe binary relations. If R is a binary relation with domain D × D , then it can be represented as a labeled graph G = ( D , E ), where E = {( x , y ) : xRy }.

示例0.17

Example 0.17

下面所示的有向图表示例 0.10 中的关系。

The directed graph shown below represents the relation in Example 0.10.

图 0.18

关系节拍图

Figure 0.18

Graph for the relation beats

文字和语言

Words and languages

单词或字符串是计算机科学的基本组成部分。定义单词的字母表可能会根据用途而变化。出于我们的目的，我们将字母表定义为非空有限集。该集合的元素是字母符号。通常，我们将使用大写希腊字母S和G表示字母表，使用小写等宽字体表示字母符号。下面显示了字母表的一些示例。

Words, or sequences of characters, are the fundamental building blocks of computer science. The alphabet over which words are defined can change depending on the application. For our purposes, we define an alphabet as a nonempty finite set. The elements of this set are the symbols of the alphabet. We will usually use the uppercase Greek letters S and G to denote alphabets and lowercase monospaced fonts for the symbols of the alphabet. Several examples of alphabets are shown below.

Σ ₁ = { 0 , 1 }

Σ ₁ = { 0 , 1 }

Σ ₂ = { a , b , c , d , e , f , g , h , i , j , k , l , m , n , o , p , q , r , s , t , u , v , w , x , y , z }

Σ ₂ = { a , b , c , d , e , f , g , h , i , j , k , l , m , n , o , p , q , r , s , t , u , v , w , x , y , z }

γ = { 0 , 1 , x , y , z }

Γ = { 0 , 1 , x , y , z }

字母表上的单词（也是字母表上的字符串）是该字母表中符号的有限序列，通常一个接一个地书写，不带空格或逗号。如果Σ ₁ = { 0 , 1 }，则01001就是Σ ₁上方的字。如果Σ ₂ = { a , b , c , … , z }，则abracadabra就是Σ ₂上方的单词。如果w是Σ上方的一个字，则w 的长度 写为 | w |, 是它包含的符号数。长度为零的字称为空字，写为ε。空白这个词在数字系统中的作用类似于0。如果单词w的长度为n，我们可以将其写为w = w ₁ w ₂ … ·w _n，其中每个w _i ∈ Σ。单词w的反转，写为w R ，是通过将w逆序书写而获得的单词（即w _n w _{n -1} ...·w ₁）。单词with是单词w的连接子串，则它是 w的子词。例如， cad是abracadabra的子词。

A word over an alphabet (also a sequence over an alphabet ) is a finite sequence of symbols from that alphabet, usually written out one after the other without spaces or commas. If Σ ₁ = { 0 , 1 }, then 01001 is a word over Σ ₁ . If Σ ₂ = { a , b , c , … , z }, then abracadabra is a word over Σ ₂ . If w is a word over Σ , the length of w , written as | w | , is the number of symbols in it. A word of length zero is called the empty word and is written as ε . The empty word plays a role analogous to 0 in the numeral system. If a word w has length n , then we can write it as w = w ₁ w ₂ … ·w _n , where each w _i ∈ Σ . The inversion of a word w , written as w R , is the word obtained from w by writing it in reverse order (that is, w _n w _{n −1} …·w ₁ ). A word z is a subword of w if it is a consistent subsequence of w . For example , cad is a subword of abracadabra .

如果我们有一个长度为m的单词x和一个长度为n的单词y，那么x和y的串联，写作xy ，就是将单词y附加到单词x的末尾得到的单词，即单词x ₁ ... x _m y ₁ ... y _n。在一个单词与其自身多重连接的情况下，我们使用指数符号x ^k，意思是

If we have a word x of length m and a word y of length n , then the concatenation of x and y , written xy , is the word obtained by appending the word y to the end of the word x , that is, the word x ₁ … x _m y ₁ … y _n . In the case of multiple concatenation of a word with itself, we use the exponential notation x ^k , meaning

k

k

xx … x。

xx … x .

单词的词典顺序与众所周知的词典顺序相同。我们偶尔会使用修改后的词典顺序，称为Shortlex，它与词典顺序相同，只是较短的单词在较长的单词之前。因此，字母表上方所有单词的短词顺序为 { 0 , 1 }

The lexicographical order of words is identical to the well-known dictionary order. We will occasionally use a modified lexicographical order, called shortlex , which is identical to the lexicographical order except that shorter words precede longer ones. Thus, the shortlex order of all words over the alphabet { 0 , 1 } is

（ε 、0、1、00、01、10、11、000、... ）。​​​​​​​

(ε, 0 , 1 , 00 , 01 , 10 , 11 , 000 , …).

如果存在一个单词z使得xz = y ，我们就说单词x是单词y的前缀，并且如果另外x ≠ y ，则x被称为单词y的真前缀。

We say that a word x is a prefix of a word y if there is a word z such that xz = y , and x is called a proper prefix of y if additionally x ≠ y .

语言是一组单词。如果一种语言中没有元素（单词）是另一个元素的正确前缀，则该语言被称为有前缀的。

A language is a set of words. A language is called prefixed if no element (word) of the language is a proper prefix of another element.

布尔逻辑

Boolean logic

布尔逻辑是围绕 TRUE 和 FALSE 这两个值构建的数学系统。尽管最初是作为纯粹的数学概念而发明的，但该系统现在被视为数字电子和计算机设计的基础。 TRUE 和 FALSE 值称为布尔（逻辑）值，通常表示为数值 1 和 0。我们在只有两种选择的情况下使用布尔值，例如高或低电压水平，一个可以是真或假的陈述，或者一个答案必须是或否的问题。

Boolean logic is a mathematical system built around the two values ​​TRUE and FALSE. Although originally invented as a purely mathematical concept, the system is now seen as the foundation for the design of digital electronics and computers. The values ​​TRUE and FALSE are calledBoolean values ​​and are often represented as the numeric values ​​1 and 0. We use Boolean values ​​in situations where only two possibilities are available, such as a high or low voltage level, a statement that can be true or false, or a question that must be answered yes or no.

布尔值的转换是通过布尔（逻辑）运算来完成的。最简单的布尔运算是求反，即NOT运算，用符号 Ø 表示。布尔值的负数是它的相反值。因此， Ø0 = 1 且 Ø1 = 0。合取，即AND运算，用符号∧标记。如果两个布尔值的值都是 1，则两个布尔值的合取为 1（真）。我们用符号∨表示替代或OR运算。如果两个布尔值中的任何一个为 1，则两个布尔值的替代值为 1。此信息可以总结如下：

The transformation of boolean values ​​is done by boolean ( logical ) operations. The simplest boolean operation is negation , or the NOT operation , denoted by the symbol ¬. The negation of a boolean value is its opposite value. Thus, ¬0 = 1 and ¬1 = 0. Conjunction , or the AND operation , is denoted by the symbol ∧ . The conjunction of two boolean values ​​is equal to 1 (true), if both of these values ​​are 1. The disjunction or OR operation is denoted by the symbol ∨ . The disjunction of two boolean values ​​is equal to 1 if either of these values ​​is equal to 1. This information can be summarized as follows:

0 ∧ 0 = 0 0 ∧ 0 = 0	0 ∨ 0 = 0 0 ∨ 0 = 0	Ø0 = 1 ¬0 = 1
0 ∧ 1 = 0 0 ∧ 1 = 0	0 ∨ 1 = 1 0 ∨ 1 = 1	Ø1 = 0 ¬1 = 0
1 ∧ 0 = 0 1 ∧ 0 = 0	1 ∨ 0 = 1 1 ∨ 0 = 1
1 ∧ 1 = 1 1 ∧ 1 = 1	1 ∨ 1 = 1 1 ∨ 1 = 1

布尔运算用于将简单的语句组合成更复杂的布尔表达式，就像我们使用+和×算术运算来构造复杂的算术表达式一样。例如，如果P是一个布尔值，代表语句“阳光灿烂”，Q代表语句“今天是星期一”，那么我们可以写P  ∧  Q来代表语句“”阳光明媚，今天是星期一”；P  ∨  Q也存在类似的关系，其中“i”被“or”替代。P和Q的值称为运算的操作数。

Boolean operations are used to combine simple statements into more complex Boolean expressions, much as we use the arithmetic operations + and × to construct complex arithmetic expressions. For example, if P is a Boolean value representing the truth value of the statement “The sun is shining” and Q represents the truth value of the statement “Today is Monday”, then we can write P  ∧  Q to represent the truth value of the statement “The sun is shining and today is Monday”; a similar relationship holds for P  ∨  Q , with “and” replaced by “or”. The values ​​P and Q are called the operands of the operation.

其他一些布尔运算会不时出现。异或( XOR ) 由符号⊕表示，如果一个或另一个操作数的值为 1，但不是两个都为 1，则其值为1 。当两个操作数具有相同的值时，等价运算用符号↔编写，其值为 1。最后，用符号 → 表示的蕴涵运算，如果其第一个（左）操作数为 1 并且第二个操作数为 0，则其值为 0；在其他情况下，→ 的值为 1。我们将此信息总结如下。

A few other Boolean operations will pop up from time to time. Exclusive or , or XOR , is denoted by the symbol ⊕ and has the value 1 if either one or the other operand is 1, but not both. The equivalence operation , denoted by the symbol ↔ , has the value 1 if both operands have the same value. Finally, the implication operation , denoted by the symbol → , has the value 0 if its first (left) operand is 1 and its second is 0; otherwise, → has the value 1. We summarize this information as follows.

0 ⊕ 0 = 0 0 ⊕ 0 = 0	0 ↔ 0 = 1 0 ↔ 0 = 1	0 → 0 = 1 0 → 0 = 1
0 ⊕ 1 = 1 0 ⊕ 1 = 1	0 ↔ 1 = 0 0 ↔ 1 = 0	0 → 1 = 1 0 → 1 = 1
1 ⊕ 0 = 1 1 ⊕ 0 = 1	1 ↔ 0 = 0 1 ↔ 0 = 0	1 → 0 = 0 1 → 0 = 0
1 ⊕ 1 = 0 1 ⊕ 1 = 0	1 ↔ 1 = 1 1 ↔ 1 = 1	1 → 1 = 1 1 → 1 = 1

这些操作之间存在各种依赖关系。事实上，我们只能用 AND 和 NOT 运算来表达所有布尔运算，如下所示。每行中的两个表达式是等效的。每行在左列中显示使用上述概念的操作记录，在右列中显示使用 AND 和 NOT 的等效操作。

There are various dependencies between these operations. In fact, we can express all Boolean operations only by the AND and NOT operations, as the following identities show. The two expressions in each row are equivalent. Each row presents in the left column the notation of the operation using the concepts discussed above, and in the right column the equivalent operations using AND and NOT.

P∨Q ​ ​ P ∨ Q	← (← P∧ ← Q ) ¬(¬ P ∧ ¬ Q )
Q → Q P → Q	← P ∨ Q ¬ P ∨ Q
P ↔ Q P ↔ Q	( P → Q ) ∧ ( Q → P ) ( P → Q ) ∧ ( Q → P )
P ⊕ Q P ⊕ Q	←( P ↔ Q ) ¬( P ↔ Q )

合取和析取的分配律在转换布尔表达式时很有用。它类似于乘除加法的分配律，即a × ( b + c ) = ( a × b )+( a × c )。布尔版本有两种形式：

The distributive law of conjunction and disjunction is useful when transforming Boolean expressions. It is similar to the distributive law of multiplication over addition, which states that a × ( b + c ) = ( a × b )+( a × c ). The Boolean version comes in two forms:

• P ∧ ( Q ∨ R ) 等于 ( P ∧ Q ) ∨ ( P ∧ R )，并且
• P ∧ ( Q ∨ R ) is equal to ( P ∧ Q ) ∨ ( P ∧ R ), and
• P ∨ ( Q ∧ R ) 等于 ( P ∨ Q ) ∧ ( P ∨ R )。
• P ∨ ( Q ∧ R ) is equal to ( P ∨ Q ) ∧ ( P ∨ R ).

数学术语总结

Summary of Mathematical Terms

字母表 – 称为符号的有限、非空对象集

Alphabet – a finite, non-empty set of objects called symbols

替代方案 – 布尔 OR 运算

Alternative – Boolean operation OR

Argument – 函数的输入（数据）值

Argument – ​​input value (data) of the function

字符串——对象的有序列表

String - an ordered list of objects

循环 – 在同一顶点开始和结束的路径

Cycle – a path that starts and ends at the same vertex

补——返回不属于给定集合的所有元素的集合的操作

Complement – ​​an operation that returns a set of all elements that do not belong to a given set

树——不包含简单循环的连通图

Tree – a connected graph that does not contain any simple cycles.

双参数关系——其域是有序对集合的关系

A binary relation is a relation whose domain is the set of ordered pairs.

域——函数所有可能的输入值（参数）的集合

Domain – the set of all possible input values ​​(arguments) of a function

元素——属于集合的对象

Element – ​​an object belonging to a set

函数——将输入值转换为输出值的操作

Function – an operation that transforms input values ​​into output values

图形 – 一组点和连接这些点对的线

Graph – a set of points and lines connecting some pairs of these points.

笛卡尔积 – 集合上的运算，创建由适当集合的元素组成的所有元组的集合

Cartesian product – an operation on sets that creates the set of all tuples composed of elements of the corresponding sets.

语言——特定字母表上的一组单词

Language – a collection of words over a certain alphabet

k -tuple – k 个对象的序列

k -tuple – a sequence of k objects

连接 – 布尔 AND 运算

Conjunction – Boolean operation AND

连接——组合单词的操作

Concatenation – the operation of joining words together

边 – 连接图顶点的线

Edge – a line connecting the vertices of a graph

无序对——包含两个元素的集合

An unordered pair is a set with two elements.

布尔运算——布尔值的运算

Boolean operation – operation on boolean values

谓词 – 子域为集合 {TRUE, FALSE} 的函数

Predicate – a function whose codomain is the set {TRUE, FALSE}

直线路径——没有重复顶点的路径

Simple path - a path without repeated vertices

反域——函数输出值的集合

Codomain – the set from which the output values ​​of the function come from

交集（集合的乘积）——集合上的运算，创建这些集合共有的一组元素

Intersection (product of sets) – an operation on sets that creates a set of elements common to these sets

等价关系——自反、对称和传递的双参数关系

Equivalence relation – a binary relation that is reflexive, symmetric, and transitive

关系——一个谓词，其域是一组k元组

Relation – a predicate whose domain is a set of k -tuples

单例集——单元素集

Singleton set – a set of one element

有向图 – 连接选定的这些点对的一组点和箭头（有序对）

Directed graph – a set of points and arrows (ordered pairs) connecting selected pairs of these points

连通图——包含连接任意两个顶点的路径的图

A connected graph is a graph containing paths connecting any two vertices.

集合并集——集合的一种运算，将参与集合的所有元素组合成一个集合

Union of sets – a set operation that combines all the elements of the participating sets into one set.

符号——字母表的一个元素

Symbol - an element of the alphabet

空词——长度为零的词

Empty word - a word of length zero

单词——特定字母表中符号的有限列表

Word – a finite list of symbols from a certain alphabet

路径 – 由边连接的图顶点序列

Path – a sequence of graph vertices connected by edges.

有序对 – 包含两个元素的列表，一个二元素字符串

Ordered pair – a list of two elements, a sequence of two elements

布尔值 - TRUE 或 FALSE 值，通常用 1 或 0 表示

Boolean value – TRUE or FALSE values, often represented by 1 or 0

顶点——图中的一个点

Vertex – a point in a graph

节点（或顶点）——图中的一个点

Node (also known as vertex) – a point in a graph

属性——也称为谓词

Property - otherwise known as predicate

空集——没有元素的集合

Empty set – a set that has no elements.

集合——一组对象

Set – a group of objects

0.3

定义、定理和证明

0.3

Definitions, Theorems and Proofs

定理和证明是数学的身体和血液，定义是数学的灵魂。这三个实体构成了数学每个分支的基础，包括我们的数学分支。

Theorems and proofs are the body and blood of mathematics, and definitions are its soul. These three entities form the basis of every branch of mathematics, including our own.

定义描述了我们使用的对象和概念。该定义可以很简单，如本章前面介绍的集合的定义，也可以很复杂，如密码系统中安全性的定义。任何数学定义的基本特征都是精确性。当我们定义一个对象时，我们必须确保清楚什么定义了该对象，什么不定义该对象。

Definitions describe the objects and concepts we use. A definition can be simple, as in the case of the definition of a set presented earlier in this chapter, or complex, as in the definition of security in a cryptographic system. The essential feature of any mathematical definition is precision. When we define an object, we must ensure that we are unambiguous about what the object defines and what it does not.

在定义了各种对象和概念之后，我们通常会制定关于它们的数学句子 （陈述）。典型的说法是某个对象具有某种属性。陈述可能真实，也可能不真实；然而，就像定义一样，它必须是精确的。不允许对其含义存在任何不确定性。

After defining various objects and concepts, we usually formulate mathematical sentences (statements) about them. A typical statement is that a certain object has a certain property. The statement may be true or false; but again like a definition, it must be precise. No uncertainty about its meaning is permitted.

证明是证明特定陈述正确的令人信服的逻辑论证。在数学中，这个论证必须是严格的。换句话说，它必须在绝对意义上具有说服力。在日常生活或法律中，证据的标准较低。在谋杀案审判中，需要“排除合理怀疑”的证据。证据的分量（证词或重要事实）可以说服陪审团认定被告无罪或有罪。然而，这样的证据在数学证明中不起任何作用。数学家要求毫无疑问的证明。

A proof is a convincing logical argument that a given statement is true. In mathematics, the argument must be rigorous; in other words, it must be convincing in an absolute sense. In everyday life or in law, the standard of proof is lower. In a murder trial, proof "beyond a reasonable doubt" is required. The weight of the evidence (or testimony or material fact) can persuade a jury to find the accused innocent or guilty. But such evidence plays no role in a mathematical proof. The mathematician demands proof beyond a reasonable doubt.

定理是已被证明正确的数学陈述。一般来说，我们保留在具有特殊含义的句子中使用这个词。我们有时会证明一些有趣的陈述，只是因为它们有助于证明其他更重要的陈述。这些类型的句子称为引理。有时，一个陈述或其证明可以很容易地推断出其他相关陈述是正确的。这样的命题称为定理的结论。

A theorem is a mathematical statement that has been proven to be true. In general, we reserve the use of this word for statements of special importance. Occasionally, we prove statements that are interesting only because they help us prove other, more significant statements. Such statements are called lemmas . Sometimes it happens that a theorem or its proof allows us to easily deduce that other related statements are true. Such statements are called corollaries of the theorem.

寻找证据

Finding evidence

确定数学陈述真假的唯一方法是进行数学证明。不幸的是，找到证据并不总是那么容易。这项任务不能简化为一组简单的规则或程序。在阅读本手册时，读者经常需要提供各种定理的证明。不过，无需提前担心！虽然没有人找到创建证据的通用公式，但有一些通用策略可以提供帮助。

The only way to determine whether a mathematical statement is true or false is to perform a mathematical proof. Unfortunately, finding a proof is not always easy. The task cannot be reduced to a simple set of rules or procedures. In the course of working through this manual, the reader will often have to provide proofs of various theorems. But do not worry in advance! Although no one has found a universal recipe for creating proofs, there are several general strategies that can help.

你应该仔细阅读你想要证明的陈述。我们真的理解所有的符号吗？值得用你自己的话重写这句话。将它们分成几部分并分别考虑。

You should carefully read the sentence you want to prove. Do you really understand the entire notation? It's worth rewriting the sentence in your own words. Divide it into parts and consider each of them separately.

有时复合句的某些部分并不是立即显而易见的。一种常见的复合句的形式是“ P if and only if Q ”，其中P和Q是数学句子。该符号是由两部分组成的句子的缩写。第一部分是“ P only if Q ”，意思是：如果P为真，则Q也为真，我们写为P⇒Q 。 第二部分是“ P if Q ”，意思是：如果Q为真，则P也为真，写为PÜQ 。这些部分中的第一部分是朝着原始陈述的一个方向，第二部分是朝着另一个方向。 “P 当且仅当Q ”写为P ⇔ Q。为了证明这种形式的定理，需要在两个方向上进行证明。通常这些方向之一比另一个更容易证明。

Sometimes the parts of a compound sentence are not immediately obvious. One common compound sentence is of the form " P if and only if Q ," where P and Q are mathematical sentences. This notation is shorthand for a two-part sentence. The first part is " P only if Q ," meaning: if P is true, then Q is true, written as P ⇒ Q . The second part is " P , if Q ," meaning: if Q is true, then P is true, written as P Ü Q . The first part of this statement is in one direction of the original statement, while the second part is in the other direction . "P if and only if Q " is written as P ⇔ Q . To prove a statement of this form, it is necessary to prove it in both directions. Often, one of these directions is easier to prove than the other.

另一种类型的复杂句子是陈述两个集合A和 B相等。第一部分声明A是B的子集，第二部分声明B是A的子集。因此，证明A = B 的一种常见方法是证明A 的每个元素也是B的元素，然后B的每个元素也是A的元素。

Another type of compound proposition is the statement that two sets A and  B are equal. The first part states that A is a subset of B , and the second part states that B is a subset of A . Hence, one common way of proving that A = B is to show that every element of A is also an element of B , and then that every element of B is also an element of A .

最后，当试图证明一个命题或命题的一部分时，必须尝试获得一种直观的“内在”感觉，了解为什么它应该是正确的。尝试示例在这里特别有帮助。如果一个命题指出某种类型的所有对象都具有某种属性，那么我们可以选择该类型的多个对象并检查它们是否确实具有该属性。下一步，你可以尝试找到一个不满足这个条件的对象，这称为反例。如果这种说法确实正确，我们将找不到反例。尝试寻找反例时看到困难出现在哪里可以帮助您理解为什么这个陈述是正确的。

Finally, when we try to prove a statement, or part of a statement, we try to get an intuitive “inner” sense of why it should be true. Experimenting with examples can be particularly helpful here. If the statement states that all objects of a certain type have a certain property, we can pick a few objects of that type and see if they really do have that property. The next step is to try to find an object that doesn’t satisfy that condition, which we call a counterexample . If the statement is really true, we won’t be able to find a counterexample. Seeing where the difficulties in trying to find a counterexample arise can help us understand why the statement is true.

示例 0.19

Example 0.19

假设我们要证明对于任意图G，该图所有顶点的度数之和为偶数。

Suppose we want to prove the statement that for any graph G, the sum of the degrees of all the vertices of that graph is an even number .

首先，我们选择一些图表并在实际中测试此语句。下面是两个例子：

First, we select a few graphs and test this statement in action. Here are two examples:

然后我们尝试寻找一个反例；换句话说，我们正在寻找一个图，其中该总和为奇数。

Then we try to find a counterexample; in other words, we look for a graph in which this sum is an odd number.

我们是否已经看出这个说法是正确的，我们如何证明它？

Can you already see that this statement is true, and how can you prove it?

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如果我们不断地试图证明一个陈述，那么值得尝试一些更简单的事情。人们可以尝试证明该陈述的一种特殊情况。例如，如果我们试图证明某个属性对于所有k > 0 都成立，我们首先尝试在k = 1时证明它。如果成功，我们会尝试k = 2，依此类推，直到我们捕获更多一般情况。如果所选择的特例难以证明，则可以尝试另一个特例，甚至可能是该特例的一个特例。

If we get stuck trying to prove a statement, it's worth trying something easier. We can try to prove a special case of the statement. For example, if we're trying to prove that some property is true for all k > 0, we first try to prove it for k = 1. If that works, we try for k = 2, and so on, until we get a more general case. If the chosen special case turns out to be hard to prove, we can try another special case, or perhaps even a special case of that special case.

最后，当我们确信已经找到证据时，我们需要适当地记录下来。一个写得很好的证明是一系列句子，每个句子都是从前面的句子中进行简单推理得出的。仔细写下证明对于帮助读者理解并确保其没有错误非常重要。

Finally, when we are sure that we have found a proof, we need to write it down properly. A well-written proof is a series of statements, each of which follows by simple reasoning from the previous statements. Writing a proof carefully is important both because it helps the reader understand it and because it helps us to make sure that it is free from errors.

以下是一些可以帮助您创建良好证据的提示。

Here are some tips to help you create good evidence.

• 要有耐心。寻找证据需要时间。如果您没有立即明白如何证明某件事，请不要担心。研究人员有时会花数周甚至数年的时间研究单个证据。
• Be patient . Finding evidence takes time. If you don't immediately see how to prove something, don't worry. Researchers sometimes work for weeks or even years on a single piece of evidence.
• 一段时间后回来继续任务。看看你想要证明的陈述，想一想，然后离开，几分钟或几小时后再回来。让你头脑中无意识的直觉部分有机会发挥作用。
• Return to the task after some time . Look at the statement you want to prove, think about it for a while, leave it, then return a few minutes or hours later. Give the unconscious intuitive part of your mind a chance to work.
• 保持整洁。当你对想要证明的命题建立直觉时，请使用简单、清晰的绘图和注释。你想要深入了解问题的本质，而马虎会阻碍这一点。此外，当你编写一个解决方案供其他人阅读时，整洁也会帮助他们理解它。
• Be neat . When you're building an intuition about a proposition you want to prove, use simple, clear drawings and notes. You want to create insight into the nature of the problem, and sloppiness gets in the way of that. Moreover, when you're writing a solution for someone else to read, neatness will help that person understand it, too.
• 保持简洁。这可以帮助您表达重要的想法，而不会陷入细节之中。良好的数学符号可以帮助您简洁地表达想法。但是，您需要确保在编写证明时，足够详细地表达您的推理，以便读者能够快速理解您想要说的内容。
• Be concise . This helps you express important ideas without getting bogged down in details. Good mathematical notation helps you to express ideas concisely. However, you need to take care when writing a proof to present your reasoning in enough detail that the reader can quickly understand what you are trying to say.

作为练习，我们将证明德摩根定律之一。

As an exercise, we will prove one of DeMorgan's laws.

定理 0.20

Theorem 0.20

对于任意两个集合A和 B，A ∪ B = A ∩ B。

For any two sets A and  B , A ∪ B = A ∩ B .

首先，这句话的意思清楚吗？如果您不明白∪或∩符号或顶部破折号的含义，请参阅第 4 页的讨论。

First, is the meaning of this statement clear? If you don't understand the meaning of the symbols ∪ or ∩ , or the dash at the top, see the discussion on page 4.

为了证明这个定理，我们需要证明集合A ∪ B和A ∩ B相等。回想一下，我们可以通过证明一个集合的每个元素也是另一个集合的元素来证明集合的相等性，反之亦然。在阅读下面的证明之前，请考虑几个例子，然后尝试自己进行证明。

To prove this theorem, we need to show that the sets A ∪ B and A ∩ B are equal. Recall that we can prove set equality by showing that every element of one set is also an element of the other set, and vice versa. Before reading the proof that follows, it is worth considering a few examples and then trying the proof yourself.

证明该定理指出两个集合A ∪ B和A ∩ B相等。我们将通过证明一个集合中的每个元素也是另一个集合中的元素来证明这一论点，反之亦然。

Proof This theorem states that two sets, A ∪ B and A ∩ B , are equal. We prove this theorem by showing that every element of one of the sets is also an element of the other, and vice versa.

假设x是集合A È B的一个元素。根据集合补集 的定义，x不属于A ∪ B。因此，根据集合和的定义，x既不是A的元素，也不是B的元素。换句话说，x属于A，  x属于B。因此，根据集合交集的定义，我们得到x属于A∩B 。

Suppose x is an element of the set A È B . By the definition of the complement of sets, it follows that x is not an element of A ∪ B . Therefore, by the definition of the union of sets, x is neither an element of A nor an element of B . In other words, x belongs to A and  x belongs to B . Hence, by the definition of set intersection, we have that x belongs to A ∩ B .

为了反方向证明该定理，假设x属于A ∩ B。这意味着x既属于A又属于B。由此可见x不属于A且x不属于B，因此不属于这些集合的和。因此x属于这些集合之和的补集；换句话说，x属于A∪B ，这就完成了定理的证明。  

To prove the theorem in the opposite direction, suppose x belongs to A ∩ B . This means that x belongs to both A and B . It follows that x does not belong to A and that x does not belong to B , and hence does not belong to the union of these sets. Hence x belongs to the complement of the union of these sets; in other words, x belongs to A ∪ B , which completes the proof of the theorem.  

现在让我们证明例 0.19 中的陈述。

Let us now prove the statement from Example 0.19.

定理 0.21

Theorem 0.21

对于任意图G，G中所有顶点的度数之和为偶数。

For any graph G, the sum of the degrees of all the vertices of G is an even number.

证明G 中的每条边都连接到两个顶点。每条边为其所连接的每个顶点的度贡献 1。因此，每条边将所有节点的度数之和加 2。因此，如果G包含e条边，则G所有顶点的度数之和为 2e，为偶数。

Proof Each edge in G is connected to two vertices. Each edge contributes 1 to the degree of each vertex to which it is connected. Thus, each edge contributes 2 to the sum of the degrees of all the nodes. Therefore, if G contains e edges, then the sum of the degrees of all the vertices of G is 2e , which is an even number.

0.4

证据类型

0.4

Types of evidence

数学证明中经常出现几种类型的论证。在这里，我们将描述计算理论中经常出现的一些内容。请注意，一个证明可以包含不止一种类型的论证，因为一个证明可以有多个不同的子证据。

Several types of arguments often appear in mathematical proofs. Here we describe several that are common in the theory of computation. Note that a proof may contain more than one type of argument, since a proof may have several different subproofs.

构造证明

Proof by construction

许多主张声称存在某种类型的物体。证明该定理的一种方法是演示如何构造这样的对象。这种技术称为构造证明（constructionproof）。

Many theorems state that a certain type of object exists. One way to prove such a theorem is to demonstrate how to construct such an object. This technique is called proof by construction .

让我们尝试使用构造证明来证明以下定理。如果图的每个顶点都有度k ，我们将图定义为k-正则图。

Let us try to use proof by construction to prove the following theorem. A graph is defined as k-regular if every vertex of the graph has degree k .

定理 0.22

Theorem 0.22

对于每个大于 2 的偶数n，都有一个具有n 个顶点的 3-正则图。

For every even number n greater than 2, there is a 3-regular graph with n vertices.

证明令n为大于2的偶数。我们根据以下过程构造一个具有n个 顶点的图G = ( V , E ) ：顶点集G为V = {0, 1, …, n − 1} ，并且边集G被设置

Proof Let n be an even number greater than 2. We construct a graph G = ( V , E ) with n  vertices according to the following procedure: the set of vertices of G is V = {0, 1, …, n − 1}, and the set of edges of G is the set

E = { { i , i + 1} ：对于 0 ≤ i ≤ n − 2} ∪ { { n − 1, 0} } ∪ { { i , i + n /2} ：对于 0 ≤ i ≤ n /2 − 1}。

E = { { i , i + 1} : for 0 ≤ i ≤ n − 2} ∪ { { n − 1, 0} } ∪ { { i , i + n /2} : for 0 ≤ i ≤ n /2 − 1}.

该图的顶点可以通过将它们均匀分布在圆上来表示。在这种情况下，由集合E的定义的顶线描述的边连接圆周围的相邻顶点对。集合E定义的底线中描述的边在圆另一侧的顶点之间延伸。这个心理图像清楚地表明G的每个顶点的度数都是 3。

The vertices of this graph can be represented by distributing them evenly on a circle. In this case, the edges described by the top line of the definition of set E connect adjacent pairs of vertices around the circle. The edges described by the bottom line of the definition of set E run between vertices on the opposite side of the circle. This mental picture clearly shows that each vertex of G has degree 3.

间接证明（通过减少矛盾）

Proof by contradiction (reduction to contradiction)

证明定理的另一种常见论证形式包括假设该定理是错误的，然后表明该假设显然会导致错误的结论，称为矛盾。我们在日常生活中经常使用这种推理，如下例所示。

Another common form of argument in proving a theorem involves assuming that the theorem is false and then showing that this assumption leads to an obviously false conclusion, called a contradiction. We use this type of reasoning frequently in everyday life, as in the following example.

示例 0.23

Example 0.23

杰克看到吉尔刚才还在外面。她发现自己没有被淋湿，所以她知道没有下雨。他证明没有下雨的“证据”是，如果下雨（该断言是错误的），吉尔就会被淋湿（显然是错误的效果）。因此不能下雨。

Jack sees Jill, who has just been outside. He notices that she is not wet, so he knows that it is not raining. His "proof" that it is not raining is that if it were raining (the false assumption that the statement is false), Jill would get wet (an obviously false consequence). Therefore, it cannot be raining.

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现在我们尝试间接证明2的平方根是一个无理数。如果一个数是分数m / n ，则该数是有理数，其中m和n是整数；换句话说，有理数是整数m和n的商。例如，2/3显然是一个有理数。如果一个数不是有理数，则该数称为无理数。

Now let's try to prove indirectly that the square root of 2 is an irrational number. A number is rational if it is a fraction m / n , where m and n are integers; in other words, a rational number is the quotient of integers m and n . For example, 2/3 is obviously a rational number. A number is called irrational if it is not rational.

定理 0.24

Theorem 0.24

√2是无理数。

√ 2 is an irrational number.

证明 为了证明的目的，我们隐含地假设√ 2是有理数。意思是

Proof For the purpose of the proof indirectly, we assume that √ 2 is a rational number. Therefore,

其中m和n是整数。如果数字m和n都可以被同一个大于 1 的整数整除，则我们将它们除以最大的此类数字（减少分数）。此操作不会更改分数的值。因此，数字m和n中至少有一个必须是奇数。

where m and n are integers. If both numbers m and n are divisible by the same integer greater than 1, we divide both by the largest such number (we reduce the fraction). This operation does not change the value of the fraction. As a result, at least one of the numbers m and n must be an odd number.

我们将方程两边同时乘以n，得到

We multiply both sides of the equation by n and get

n √ 2 = m。

n √ 2 = m .

我们将两边平方并得到

We square both sides and get

2 n ² = m ²。

2 n ² = m ² .

由于m ²是整数n ^{2 的}两倍，所以我们知道m ²是偶数。因此m本身也是偶数，因为奇数的平方总是奇数。因此我们可以写成m = 2 k，其中k是某个整数。用2 k代替m，我们得到

Since m ² is a double of the integer n ² , we know that m ² is an even number. Therefore m itself is also even, since the square of an odd number is always odd. We can therefore write m = 2 k , where k is some integer. Substituting 2 k for m , we get

2 n ² = (2 k ) ²

2n2 = ( ^2k ) ²​

= 4k ²。

= 4k2 .^​

两边同时除以 2，我们得到

Dividing both sides by 2, we get

n ² = 2 k ²。

n ² = 2 k ² .

然而，该结果表明n ²是偶数，因此n是偶数。因此我们证明m和n都是偶数。然而，我们之前已经减少了m和n（减少了分数），因此两者不能同时为偶数 - 这是一个矛盾。

However, this result shows that n ² is even, and therefore n is even. We have therefore shown that both m and n are even. However, we have previously reduced m and n (reduced the fraction), so both cannot be even at the same time – a contradiction.

感应证明

Inductive proof

归纳证明是一种先进的方法，用于证明某个无限集合的所有元素都具有特定的属性。例如，我们可以使用归纳证明来证明某个算术表达式对于每个变量赋值都计算出所需的值，或者程序在所有步骤或所有输入值中都能正确运行。

An inductive proof is an advanced method for showing that all elements of an infinite set have a certain property. For example, we can use an inductive proof to show that a certain arithmetic expression evaluates to the desired value for every variable assignment, or that a program executes correctly at all steps or for all input values.

为了说明归纳证明是如何工作的，让我们假设一个无限集合作为自然数集合N = {1, 2, 3, …}，我们将所研究的属性称为P。我们的目标是证明P ( k ) 对于每个自然数k都成立。换句话说，我们要证明P (1) 为真，就像P (2)、P (3)、P (4) 等等一样。

To illustrate how the inductive proof works, let us take the infinite set to be the set of natural numbers N = {1, 2, 3, …}, and call the property in question P . Our goal is to show that P ( k ) is true for every natural number k . In other words, we want to prove that P (1) is true, as are P (2), P (3), P (4), and so on.

每个归纳证明都由两部分组成，基本步骤（归纳基础）和归纳步骤。这些部分中的每一个都是一个单独的证据。在基本步骤中，我们证明P (1)为真。归纳步骤证明，对于每个i ≥ 1，如果P ( i ) 为真，则P ( i + 1)也为真。

Every inductive proof consists of two parts, a basic step ( induction basis ) and an inductive step . Each of these parts is an individual proof. In the basic step, we prove that P (1) is true. The inductive step proves that for all i ≥ 1, if P ( i ) is true, then P ( i + 1) is also true .

证明这两部分后，我们得到了期望的结果 - 具体来说，P ( i ) 对于所有i都成立。为什么？首先，我们知道P (1) 为真，因为我们在基本步骤中证明了它。其次，我们知道P (2) 为真，因为归纳步骤证明，如果P (1) 为真，则P (2) 为真，并且我们已经知道P (1) 为真。第三，我们知道P (3)为真，因为归纳步骤证明如果P (2)为真，则P (3)为真，并且我们已经知道P (2)为真。我们对所有自然数继续这个过程，证明P (4) 为真，P (5) 为真，依此类推。

After proving both parts, we get the desired result—namely, that P ( i ) is true for all i . Why? First, we know that P (1) is true, since we proved it in the basic step. Second, we know that P (2) is true, since the inductive step proves that if P (1) is true, then P (2) is true, and we already know that P (1) is true. Third, we know that P (3) is true, since the inductive step proves that if P (2) is true, then P (3) is true, and we already know that P (2) is true. We continue this process for all natural numbers, showing that P (4) is true, P (5) is true, and so on.

一旦理解了上一段，您就可以轻松应用此方法的变体和概括。例如，基本步骤不一定要从 1 开始；我们可以从b的任何值开始。在这种情况下，归纳证明将表明P ( k ) 对于任何不小于b 的k都成立。

Once the previous paragraph is understood, variations and generalizations of this method can be easily applied. For example, the basic step does not necessarily have to start at 1; we can start at any value b . In that case, an inductive proof will show that P ( k ) is true for all k not less than b .

在归纳步骤中， P ( i )为真的假设称为归纳假设（inductor suggest）。有时，使用更强的归纳假设来说明P ( j ) 对于所有j ≤ i都成立是有用的。归纳证明仍然有效，因为当我们想要证明P ( i + 1) 为真时，我们已经证明P ( j ) 对于每个j ≤ i都为真。

In the inductive step, the assumption that P ( i ) is true is called the inductive hypothesis ( inductive assumption ). It is sometimes useful to use a stronger inductive hypothesis stating that P ( j ) is true for all j ≤ i . The inductive proof will still work, since when we want to show that P ( i + 1) is true, we have already proven that P ( j ) is true for all j ≤ i .

归纳证明的书写格式如下：

The format for writing inductive proofs is as follows:

基本步骤：我们证明P (1)为真。

Basic Step : We prove that P (1) is true.

⋮

⋮

归纳步骤：对于每个i≥1，我们假设P ( i )为真，然后用这个假设来证明P ( i +1)为真。

Inductive Step : For each i ≥ 1, we assume that P ( i ) is true, and then use this assumption to show that P ( i + 1) is true.

⋮

⋮

现在我们将通过归纳证明用于计算每月抵押贷款分期付款的公式是正确的。买房时，许多人会借入部分购房所需资金，然后在一定期限内偿还债务。还款条款规定每月支付固定金额以支付利息以及部分原始金额，以便在例如 30 年内偿还全部金额。每月还款额的计算公式虽然神秘，但其实很简单。它涉及很多人的日常事务，所以应该很有趣。我们将使用归纳法来证明其正确性，这将为该技术提供一个很好的说明。

We will now prove by induction the correctness of the formula used to calculate the monthly payments on mortgages. When buying a house, many people borrow part of the money needed to make the purchase and then repay the debt over a specified number of years. The repayment terms call for a fixed amount to be paid each month to cover interest as well as part of the original amount, so that the total is paid back in, say, 30 years. The formula for calculating the monthly payments is shrouded in mystery, but it is actually quite simple. It concerns many people's daily lives, so it should be interesting. We will use induction to prove its correctness, which will provide a good illustration of the technique.

首先，我们将定义几个变量的名称和含义。令P为资本，即原始贷款的总和。设I > 0 为贷款年利率，其中I = 0.06 表示利率为 6%。令Y为每月分期付款。为了方便起见，我们将使用I来定义变量M，即每月乘数。这是您的贷款每月因应计利息而变化的利率。根据标准银行惯例，月利率为年利率的十二分之一，即M = 1 + I /12，利息按月计息（按月复利）。

First, let’s define the names and meanings of a few variables. Let P be the principal , or the amount of the original loan. Let I > 0 be the annual percentage rate of the loan, where I = 0.06 means an interest rate of 6%. Let Y be the monthly payment. For convenience, we will use I to define the variable M , or the monthly multiplier. This is the rate of change in the loan each month due to interest accrual. In accordance with standard banking practice, the monthly interest rate is one-twelfth of the annual rate, or M = 1 + I /12, and interest is compounded monthly (monthly compounding).

每个月都会发生两件事。首先，由于每月乘数，贷款规模往往会增加。其次，该金额由于每月还款而减少。令P _t为第 t个月后剩余待偿还的贷款金额。按照这个惯例，P ₀ = P 是原始贷款，P ₁  =  MP ₀  −  Y是一个月后的贷款金额，P ₂ = MP ₁ − Y是两个月后的贷款金额。现在我们准备通过t 的归纳来制定和证明一个定理，给出P _t值的公式。

Each month, two things happen. First, the size of the loan tends to increase because of the monthly multiplier. Second, the size of the loan tends to decrease because of the monthly payment. Let P _t be the size of the loan remaining to be paid after the tth month. With this convention, P ₀ = P is the original loan, P ₁  =  MP ₀  −  Y is the size of the loan after one month, and P ₂ = MP ₁ − Y is the size of the loan after two months. We are now ready to state and prove by induction on t the theorem that gives a formula for the values ​​of P _t .

定理 0.25

Theorem 0.25

对于每个t ≥ 0 都成立

For every t ≥ 0 it holds

证据

Evidence

基本步骤：我们证明该公式对于t = 0 有效。如果t = 0，则该公式变为

Basic Step : We prove that the formula is valid for t = 0. If t = 0, then the formula takes the form

右边可以通过注意到M ⁰ = 1 来简化。因此我们得到

The right-hand side can be simplified by noting that M ⁰ = 1. We thus obtain

P ₀ = P ,

P ₀ = P ,

这是正确的，因为我们已将P ₀定义为P。这样我们就证明了归纳法的基本步骤是满足的。

which is true since we defined P ₀ as P . We have thus proved that the fundamental step of induction is satisfied.

归纳步骤：对于每个k ≥ 0，我们假设如果公式对于t = k成立，那么它对于t = k + 1 成立。归纳假设指出：

Inductive step : For each k ≥ 0, we assume that if the formula holds for t = k , then it follows that it holds for t = k + 1. The inductive hypothesis states that

我们的目标是证明

Our goal is to prove that

我们将通过以下步骤来完成此任务。由于P _{k +1}是根据P _k定义的，我们知道

We will do this in the following steps. Since P _{k +1} is defined in terms of P _k , we know that

P _{k +1} = P _k M − Y。

P _{k +1} = P _k M − Y .

因此，使用归纳假设计算P _k，我们得到

Thus, using the induction hypothesis to calculate P _k , we get

乘以M并变换Y后我们得到

After multiplying by M and transforming Y we get

因此，公式对于t = k +1是正确的，从而完成了定理的证明。

Thus, the formula is valid for t = k + 1, which completes the proof of the theorem.

在问题0.15中，使用上述公式计算当前的贷款分期付款。

In task 0.15, use the above formula to calculate the current loan installments.

练习

Exercises

0.1 检查以下集合的形式表示，以了解它们包含哪些元素。对于每个集合，用日常语言写一个简短的非正式描述。

0.1 Examine the following formal representations of sets to understand what elements they contain. For each set, write a short informal description in everyday language.

{1, 3, 5, 7, … }

a. {1, 3, 5, 7, …}

b. {…,−4,−2, 0, 2, 4, …}

b. {…,−4,−2, 0, 2, 4, …}

c. { n : n = 2 m对于属于N的某个m }

c. { n : n = 2 m for some m belonging to N }

d. { n :对于属于N的某个m ， n = 2 m，对于N中的某个k ， n = 3 k }

d. { n : n = 2 m for some m belonging to N , and n = 3 k for some k from N }

e. { w : w是由 0 和 1 组成的单词，w等于反转的w }

e. { w : w is a word made up of zeros and ones and w is equal to w reversed }

f. { n : n是整数且n = n + 1}

f. { n : n is an integer and n = n + 1}

0.2 写出下列集合的正式定义。

0.2 Write formal definitions of the following sets.

a. 包含数字 1、10 和 100 的集合

a. The set containing the numbers 1, 10 and 100

b. 包含所有大于 5 的整数的集合

b. The set containing all integers greater than 5

c. 包含所有小于 5 的自然数的集合

c. The set containing all natural numbers less than 5

d. 包含单词 aba 的集合

d. A set containing the word aba

e. 包含空词的集合

e. A set containing an empty word

f. 不包含任何内容的集合

f. A set that contains nothing

0.3 设A为集合 { x , y , z } ，B为集合 { x , y } 。

0.3 Let A be the set { x , y , z}, and B be the set { x , y }.

A 是B的子集吗？

a. Is A a subset of B ?

b. B 是A的子集吗？

b. Is B a subset of A ?

C. A ∪ B 是什么？

c. What is A ∪ B ?

d. A ∩ B 是什么？

d. What is A ∩ B ?

e. 什么是A × B？

e. What is A × B ?

f. B 的幂集是多少？

f. What is the power set of set B ?

0.4 如果 A 有a 个元素，B有b 个元素，则A × B包含多少个元素？证明你的答案合理。

0.4 If A contains a elements and B has b elements, how many elements will A × B contain ? Justify your answer.

0.5 如果C是一个有c个元素的集合，那么C的幂集包含多少个元素？证明你的答案合理。

0.5 If C is a set with c elements, how many elements does the power set of C contain ? Justify your answer.

0.6 设X为集合 {1, 2, 3, 4, 5}，Y为集合 {6, 7, 8, 9, 10}。一元函数f : X → Y和二元函数g : X × Y → Y如下表所示。

0.6 Let X be the set {1, 2, 3, 4, 5} and Y be the set {6, 7, 8, 9, 10}. The unary function f : X → Y and the binary function g : X × Y → Y are described by the following tables.

n n	f ( n ) f ( n )		G g	6 6	7 7	8 8	9 9	10 10
1 1	6 6		1 1	10 10	10 10	10 10	10 10	10 10
2 2	7 7		2 2	7 7	8 8	9 9	10 10	6 6
3 3	6 6		3 3	7 7	7 7	8 8	8 8	9 9
4 4	7 7		4 4	9 9	8 8	7 7	6 6	10 10
5 5	6 6		5 5	6 6	6 6	6 6	6 6	6 6

f (2) 的值是多少 ？

a. What is the value of f (2)?

b. f 的域和子域是什么？

b. What is the domain and codomain of f ?

c. g (2, 10) 的值是多少？

c. What is the value of g (2, 10)?

d. g 的域和子域是什么？

d. What is the domain and codomain of g ?

e.g (4, f (4)) 的值是多少？

e. What is the value of g (4, f (4))?

0.7 对于每个点，给出满足给定条件的关系的示例。

0.7 For each point, give an example of a relation that satisfies the given condition.

a. 自反且对称，但不及物

a. Reflexive and symmetric, but not transitive

b. 自反和及物，但不对称

b. Reflexive and transitive, but not symmetric

c. 对称且传递，但不自反

c. Symmetric and transitive, but not reflexive

0.8 考虑一个无向图G  = ( V ,  E )，其中V是顶点集，是 {1, 2, 3, 4}，E是边集，是 {{1, 2}, {2, 3} 、{1, 3}、{2, 4}、{1, 4}}。画出图形G。每个顶点的度数是多少？在图G的图中，标记从顶点 3 到顶点 4 的路径。

0.8 Consider an undirected graph G  = ( V ,  E ), where V , the set of vertices, is {1, 2, 3, 4} and E , the set of edges, is {{1, 2}, {2, 3}, {1, 3}, {2, 4}, {1, 4}}. Draw the graph G . What is the degree of each vertex? On the drawing of G, mark the path from vertex 3 to vertex 4.

0.9 写出下图的正式描述。

0.9 Write a formal description of the graph below.

任务

Tasks

0.10 找出下面证明中 2 = 1 的错误。

0.10 Find the error in the proof below that 2 = 1.

考虑等式a = b。我们将等式两边乘以a，得到a ² = ab。我们将等式两边减去b ²，得到a ² − b ² = ab − b ²。我们将两边重新排列为 ( a + b )( a − b ) = b ( a − b )，然后将两边除以 ( a − b )，得到a + b = b。最后，我们用 1 代替a和 b，得到 2 = 1。

Consider the equation a = b . We multiply both sides of the equation by a , getting a ² = ab . We subtract b ² from both sides of the equation , getting a ² − b ² = ab − b ² . We transform both sides to the form ( a + b )( a − b ) = b ( a − b ), and then divide both sides by ( a − b ), getting a + b = b . Finally, we substitute 1 for a and  b , which gives 2 = 1.

0.11 令S ( n ) = 1 + 2 + … + n为前n 个自然数的和，令C ( n ) = 1 ³ + 2 ³ + … + n ^3为前n 个自然数的立方和自然数。使用n的归纳法证明以下等式，得出令人惊讶的结论：对于每个n ， ^C ( n ) = S2 ( n ) 。

0.11 Let S ( n ) = 1 + 2 + … + n be the sum of the first n natural numbers, and let C ( n ) = 1 ³ + 2 ³ + … + n ³ be the sum of the first n cubes of the natural numbers. Prove the following equalities using induction on n to reach the surprising conclusion that C ( n ) = S ² ( n ) for all n .

S ( n ) = ¹ / ₃ n ( n + 1) 。

a. S ( n ) = ¹ / ₃ n ( n + 1).

b. C ( n ) = ¹ / ₃ ( n ⁴ + 2 n ³ + n ² ) = ¹ / ₃ n ² ( n + 1) ²。

b. C ( n ) = ¹ / ₃ ( n ⁴ + 2 n ³ + n ² ) = ¹ / ₃ n ² ( n + 1) ² .

0.12 找出以下证明中所有马颜色相同的错误。

0.12 Find the error in the proof below that all horses are the same color.

定理：在任何一组h的马中，所有马的颜色都是相同的。

Theorem : In any set h of horses, all the horses are the same color.

证明：通过h归纳。

Proof : By induction on h .

基本步骤：对于h = 1。在任何仅包含一匹马的集合中，所有马显然都具有相同的颜色。

Basic step : For h = 1. In any set containing only one horse, all horses obviously have the same color.

归纳步骤：对于k ≥ 1，我们假设该命题对于h = k成立，并证明对于h = k +1也成立包含k +1 匹马集合H。我们将证明这组中的所有马都是相同的颜色。仅包含k匹马的集合H ₁。根据归纳假设，集合H ₁中的所有马都是相同颜色的。现在我们返回移除的马并移除另一匹以获得集合H ₂。同样的道理， H ₂中的所有马都是相同的颜色。因此，集合中的所有马必须具有相同的颜色，这就完成了证明。

Inductive Step : For k ≥ 1, we assume that the thesis is true for h = k and we will show that it is also true for h = k + 1. Take any set H containing k + 1 horses. We will show that all horses in this set are of the same color. We remove one horse from this set to obtain a set H ₁ containing only k horses. By the inductive assumption, all horses in H ₁ are of the same color. Now we return the removed horse and remove another one to obtain the set H ₂ . By the same reasoning, all horses in H ₂ are of the same color. Therefore, all horses in the set must have the same color, which completes the proof.

0.13 证明任何具有两个或多个顶点的无环图都包含两个等度的顶点。

0.13 Prove that every loop-free graph with two or more vertices contains two vertices of equal degree.

^A* 0.14 拉姆齐定理。设G是一个图。 G中的派是一个子图，其中每两个顶点由一条边连接。 Anticlique，也称为独立集，是其中任何两个顶点都没有边连接的子图。证明每个具有n 个顶点的图都包含一些或一个至少_具有^1/3 log ₂ n 个顶点的反派。

^A* 0.14 Ramsey's Theorem . Let G be a graph. A clique in G is a subgraph in which every two vertices are connected by an edge. An anticlique , also called an independent set , is a subgraph in which any two vertices are not connected by an edge. Prove that every graph with n vertices contains either some or an anticlique with at least ¹ / ₃ log ₂ n vertices.

^A 0.15利用定理0.25推导出根据贷款价值P、利率I和还款次数 t 计算每月贷款还款额的公式。假设还款t次后，债务金额减少到 0。使用此公式计算 30 年期贷款、每月还款 360 次、初始贷款价值 100,000 美元、年利率 5% 的每月还款额。

^A 0.15 Use Theorem 0.25 to derive a formula for calculating the monthly payment for a loan from the loan amount P , the interest rate I , and the number of payments  t . Assume that after making t payments, the amount owed is reduced to 0. Use this formula to calculate the monthly payment for a 30-year loan with 360 monthly payments for an initial loan amount of $100,000 at an annual interest rate of 5%.

精选解决方案

Selected solutions

0.14 为两堆顶点准备空间：A和 B。然后逐一考虑所有顶点，如果给定顶点 x 的度数大于尚未考虑的顶点的一半，则将其放入A堆，否则放入B堆；如果x被放置在A堆中（ B堆中），则从图中删除 x 未连接的所有顶点。执行这些操作直到考虑或删除所有顶点。我们在每个步骤中删除不超过一半的顶点，因此我们将至少采取 log _2n步骤。在每一步中，都会将一个顶点添加到其中一个堆中，因此其中一个堆的大小至少为¹ / ₃ log ₂ n顶点。堆 A包含形成团的顶点，堆B包含形成反团的顶点。

0.14 Prepare space for two piles of vertices: A and  B . Then consider all the vertices in turn and if a given vertex x has degree greater than half of the vertices not yet considered, then place it in pile A and otherwise in pile B ; if x was placed in pile A (in pile B ), remove from the graph all vertices to which x is not (is) connected. Perform these operations until all vertices have been considered or removed. In each step we remove no more than half of the vertices, so we take at least log ₂ n steps. In each step a vertex is added to one of the piles, so one of them has size at least ¹ / ₃ log ₂ n vertices. Pile  A contains vertices forming a clique, and pile B contains vertices forming an anticlique.

0.15 假设P _t = 0，求解Y公式，得到公式：Y = PM ^t ( M  − 1)//( M ^t − 1)。对于P = $100,000、I = 0.05 且t = 360，我们得到M = 1 + (0.05)/12。然后，您可以使用计算器确定每月付款为Y » 536.82 美元。

0.15 We set P _t = 0 and solve for Y to get the formula: Y = PM ^t ( M  − 1)//( M ^t − 1). For P = $100,000, I = 0.05, and t = 360, we get M = 1 + (0.05)/12. We can then use a calculator to determine that Y's monthly payment is $536.82.

第一部分自动化和语言

。

PART I AUTOMATS AND LANGUAGES

.

1

。正则语言

1

. Regular languages

计算理论始于这样一个问题：什么是计算机？这可能是一个愚蠢的问题，因为每个人都知道我写的是计算机。但这些真实的计算机很复杂——太复杂了，我们无法直接在它们之上构建可管理的数学理论。相反，我们将使用称为计算模型的理想化计算机。与科学中的任何模型一样，计算模型在某些方面可能是准确的，但在其他方面可能不准确。因此，我们将根据我们想要关注的特征使用许多不同的计算模型。我们将从最简单的模型开始，称为有限状态机( FSM )或有限自动机。

The Theory of Computation begins with the question: What is a computer? This is probably a silly question, since everyone knows that what I’m writing about is a computer. But real computers are complex—too complex for us to be able to construct a manageable mathematical theory directly on them. Instead, we’ll use an idealized computer called a computational model . As with any model in science, a computational model can be accurate in some respects but probably not in others. So we’ll use a variety of different computational models, depending on the features we want to focus on. We’ll start with the simplest model, called a finite-state machine ( FSM ) or finite state machine .

1.1

有限自动机

1.1

Finite automata

有限状态机对于内存极其有限的计算机来说是很好的模型。这么小的内存电脑能做什么呢？很多有用的东西！事实上，我们一直在与此类计算机打交道，因为它们构成了各种机电设备的核心。

Finite state machines are good models for computers with extremely limited amounts of memory. What can a computer with such a small amount of memory do? Lots of useful things! In fact, we deal with such computers all the time, as they are the core of various electromechanical devices.

这种设备的示例可以是自动门控制器。通常在超市入口和出口发现，当控制器检测到有人接近时，自动门就会打开。自动前门有一个垫子，可以检测即将进门的人的存在。第二个垫子位于门口的后部，以便控制器可以将门保持打开足够长的时间以供人员通过，并防止门在打开时撞到后面的人。此配置如下图所示。

An example of such a device is an automatic door controller. Commonly found in supermarket entrances and exits, automatic doors open when the controller detects a person approaching. The automatic door at the front has a pad that detects the presence of a person who is about to pass through the door. A second pad is located at the rear of the passage, so that the controller can hold the door open long enough for the person to pass through, and also to prevent the door from hitting someone standing behind it as it opens. This configuration is shown in the figure below.

图 1.1

自动门俯视图

Figure 1.1

Top view of automatic doors

控制器处于两种状态之一：“OPEN”（打开）或“CLOSED”（关闭），代表适当的门状态。如下图所示，我们有四种可能的输入条件：“FRONT”（表示该人站在过道的前垫上）、“REAR”（表示该人站在过道的后垫上） ）、“BOTH”（意味着人们站在两个垫子上）和“NEITHER”（意味着没有人站在任一垫子上）。

The controller is in one of two states: "OPEN" or "CLOSED", representing the corresponding state of the door. As you can see in the following figures, there are four possible entry conditions: "FRONT" (meaning that the person is standing on the front pad of the passage), "REAR" (meaning that the person is standing on the pad at the back of the passage), "BOTH" (meaning that people are standing on both pads), and "NEITHER" (meaning that no one is standing on either pad).

图 1.2

自动门控制器状态图

Figure 1.2

State diagram for automatic door controller

		输入信号 input signal
		两者都不 NEITHER	正面 FRONT	后部 REAR	两个都 BOTH
状态 state	关闭 CLOSED	关闭 CLOSED	打开 OPEN	关闭 CLOSED	关闭 CLOSED
	打开 OPEN	关闭 CLOSED	打开 OPEN	打开 OPEN	打开 OPEN

图 1.3

自动门控制器状态转换表

Figure 1.3

State Transition Table for Automatic Door Controller

控制器根据其接收到的输入信号从一个状态移动到另一个状态。如果它处于 CLOSED 状态并且接收到 NEITHER 或 REAR 信号，则它保持在 CLOSED 状态。此外，如果它收到两个输入，它会保持关闭状态，因为打开门意味着有撞到站在后垫上的人的风险。然而，当 FRONT 信号到达时，它进入 OPEN 状态。在OPEN状态下，如果接收到FRONT、REAR或BOTH输入信号，则保持OPEN状态。如果没有输入信号到达，控制器将返回到“关闭”状态。

The controller changes from one state to another depending on the input signal it receives. If it is in the CLOSED state and receives a NEITHER or REAR signal, it remains in the CLOSED state. Additionally, if it receives a BOTH input signal, it remains in the CLOSED state because opening the door would mean the risk of hitting someone standing on the rear pad. However, when the FRONT signal arrives, it changes to the OPEN state. In the OPEN state, if it receives a FRONT, REAR or BOTH input signal, it remains in the OPEN state. If a NEITHER input signal arrives, the controller returns to the CLOSED state.

例如，控制器可以在关闭状态下启动并接收一系列前、后、均不、前、两者、均不、后和均不输入信号。结果，会经历CLOSED（初始）、OPEN、OPEN、CLOSED、OPEN、OPEN、CLOSED、CLOSED、CLOSED等一系列状态。

For example, a controller may start in the CLOSED state and receive a series of input signals FRONT, REAR, NEITHER, FRONT, BOTH, NEITHER, REAR, and NEITHER. As a result, it will cycle through a series of states CLOSED (initial), OPEN, OPEN, CLOSED, OPEN, OPEN, CLOSED, CLOSED, and CLOSED.

将自动门控制器视为有限状态机很有用，因为它可以让您理解标准表示形式，如图 1.2 和 1.3 中的表示形式。该控制器是一台只有一位内存的计算机，能够记录控制器所处的两种状态中的哪一种。其他常见设备的驱动程序内存稍大。在电梯控制器中，状态可以代表电梯所在的楼层，并且输入数据可以是从按钮接收的信号。该计算机可能需要许多位来存储该信息。各种家用电器（例如洗碗机或电子恒温器）的控制器以及数字时钟和计算器的部件是内存有限的计算机的其他示例。在设计此类设备时，必须考虑有限自动机的方法和术语。

Considering an automatic door controller as a finite-state machine is useful because it allows us to understand standard representations such as those in Figures 1.2 and 1.3. This controller is a computer that has only a single bit of memory that can record which of two states the controller is in. Other common devices have controllers with slightly larger memories. In an elevator controller, the state might represent the floor on which the elevator is located, and the input might be signals received from buttons. This computer might need many bits to store this information. Controllers for various home appliances, such as dishwashers or electronic thermostats, and parts of digital clocks and calculators, are additional examples of computers with finite memory. Finite-state machine methodology and terminology must be kept in mind when designing such devices.

有限状态机及其概率对应物马尔可夫链在尝试识别数据模式时是有用的工具。此类设备用于语音处理和光学字符识别。马尔可夫链甚至用于建模和预测金融市场的价格变动。

Finite state machines and their probabilistic counterpart, Markov chains , are useful tools when trying to recognize patterns in data. Such devices are used in speech processing and optical character recognition. Markov chains are even used to model and predict price movements in financial markets.

现在我们将从数学的角度来看有限自动机。我们将制定有限自动机的精确定义、描述和操作有限自动机的术语以及显示其优点和局限性的理论结果。除了更好地理解有限状态机是什么以及它们能做什么和不能做什么之外，这种理论发展还将使读者能够在简单的环境中练习并更有信心地使用数学定义、定理和证明。

We will now look at finite state machines from a mathematical perspective. We will develop a precise definition of a finite state machine, a terminology for describing and manipulating finite state machines, and theoretical results that illustrate their power and limitations. In addition to providing a better understanding of what finite state machines are and what they can and cannot do, such theoretical developments will allow the reader to practice and become more confident in using mathematical definitions, theorems, and proofs in a simple environment.

首先，为了描述有限自动机的数学理论，我们将进行抽象描述，而不涉及任何具体应用。下图显示了一个名为M ₁的有限自动机。

At first, to describe the mathematical theory of finite automata, we will do it abstractly, without referring to any specific application. The figure below shows a finite automaton called M ₁ .

图 1.4具有三种状态的

有限机M ₁

Figure 1.4

Finite state machine M ₁ with three states

图1.4称为M ₁状态图 。它具有三个状态，描述为q ₁、q ₂和q ₃。初始状态q ₁由从外部进入的箭头标记。接受q _{2 的}状态是用双圆圈标记的状态。从一种状态引导到另一种状态的箭头称为转换。

Figure 1.4 is called the state diagram of M ₁ . It has three states labeled q ₁ , q ₂ , and q ₃ . The initial state q ₁ is marked with an arrow entering it from the outside. The accepting state q ₂ is the one marked with a double circle. The arrows leading from one state to another are called transitions .

当该自动机接收到输入词（例如1101 ）时，它会对其进行处理并传递结果，该结果是Accept或拒绝。目前，为了简单起见，我们将只处理这些是/否结果。处理开始于机器M ₁的初始状态。自动机从左到右一一读取输入单词中的符号。读取每个符号后，M ₁沿着转换从一种状态移动到另一种状态，并以该符号作为其标签。读取最后一个符号后，M ₁创建输出。如果M ₁当前处于接受状态，则该输出为接受，如果不是，则拒绝。

When this automaton receives an input word, for example 1101 , it processes it and produces an output, which is either accept or reject . For now, for simplicity, we will only deal with such yes/no outputs. Processing begins in the initial state of the automaton M ₁ . The automaton reads the symbols from the input word one by one, from left to right. After reading each symbol, M ₁ moves from one state to another along a transition with the given symbol as its label. After reading the last symbol, M ₁ produces an output. This output is accept if M ₁ is currently in an accepting state, or reject if it is not.

例如，当我们将输入字1101传递给图1.4中的M1机器时，处理过程如下：

For example, when we pass input word 1101 to machine M1 in Figure 1.4, processing is as follows:

1. 从状态q ₁开始。
2. Start in state q ₁ .
3. 读取1并从状态q ₁转到q ₂。
4. Read 1 and move from state q ₁ to q ₂ .
5. 读取1并从状态q ₂移动到q ₂。
6. Read 1 and move from state q ₂ to q ₂ .
7. 读取0并从状态q ₂转到q ₃。
8. Read 0 and move from state q ₂ to q ₃ .
9. 读取1并从状态q ₃移动到q ₂。
10. Read 1 and move from state q ₃ to q ₂ .
11. 接受，因为输入处理完成后M ₁处于接受状态q _{2 。}
12. Acceptance , since M ₁ is in the accepting state q ₂ after completing input processing.

对该机器进行各种输入单词的实验表明它接受单词1、01、11或0101010101。事实上，M _{1接受任何以}1结尾的单词，因为每当它读取符号1时，它就会进入接受q ₂的状态。此外，它还接受单词100、0100、110000和0101000000，以及最后一个1之后以偶数个零结尾的任何其他单词。拒绝其他单词，例如0、10、101000。我们能否描述一种由M ₁接受的所有单词组成的语言？我们稍后会这样做。

Experimenting with this machine for a variety of input words reveals that it accepts the words 1 , 01 , 11 , and 0101010101 . In fact, M ₁ accepts any word ending in 1 , since it enters the state accepting q ₂ whenever it reads the symbol 1 . In addition, it accepts the words 100 , 0100 , 110000 , and 0101000000 , as well as any other words ending with an even number of zeros following the final 1 . It rejects other words, such as 0 , 10 , 101000 . Could we describe the language consisting of all the words that M ₁ accepts ? We will do that in a moment.

有限自动机的形式定义

Formal definition of a finite state machine

在上一节中，我们使用状态图来表示有限状态机。现在我们将正式定义一个有限自动机。虽然状态图更容易直观地理解，但我们也需要正式定义有两个原因。

In the previous section, we used state diagrams to represent finite state machines. Now we define a finite state machine formally. Although state diagrams are easier to understand intuitively, there are two reasons why we also need a formal definition.

首先，形式定义是精确的。它解决了有限状态机中允许的任何不确定性。如果我们不确定有限状态机是否允许有 0 个接受状态，或者对于每个可能的输入符号是否必须从每个状态中恰好有一次转换，我们可以查看正式定义并检查答案是否为“是”两种情况。其次，形式化定义引入了符号。好的符号可以帮助您清晰地反映和表达您的想法。

First, the formal definition is precise. It resolves any uncertainty about what is allowed in a finite state machine. If we are unsure whether it is allowed for a finite state machine to have 0 accepting states, or whether it must have exactly one transition out of each state for each possible input symbol, we can look at the formal definition and see that the answer is yes in both cases. Second, the formal definition introduces notation. Good notation helps us to reason and express our thoughts clearly.

正式定义的语言有些复杂，与法律文件的语言有些相似。两者都必须准确，每个细节都必须写得没有错误。

The language of a formal definition is somewhat convoluted, bearing some resemblance to the language of legal documents. Both must be precise and every detail must be written flawlessly.

有限状态机由几个部分组成。它包含一组状态和规则，用于根据输入符号从一个状态转换到另一个状态。它有一个输入字母表，包括允许的输入符号，以及一个初始状态和一组接受状态。正式定义表明，有限自动机是这五个对象的列表：一组状态、输入字母表、转换规则、初始状态和接受状态。在数学语言中，五个元素的列表通常称为有序五元组（5 元组）。因此，我们将有限自动机定义为由这五个部分组成的元组。

A finite state machine consists of several parts. It contains a set of states and rules for transitioning from one state to another depending on an input symbol. It has an input alphabet, which includes the allowed input symbols, as well as an initial state and a set of accepting states. The formal definition says that a finite state machine is a list of these five objects: the set of states, the input alphabet, the transition rules, the initial state, and the accepting states. In mathematical language, a list of five elements is often called an ordered five (5-tuple). So we define a finite state machine as a tuple consisting of these five parts.

为了描述状态之间转换的规则，我们将使用转换函数，通常用符号δ表示。如果有限状态机包含从状态x到状态y的转换（箭头） ，并用输入符号1标记，这意味着如果机器在读取1时处于状态x，则会转换到状态y。我们可以使用传递函数表示相同的动作，即δ ( x , 1) = y。这种类型的符号是一种数学捷径。将所有这些放在一起，我们得到了有限状态机的正式定义。

To describe the rules for transitioning between states, we will use a transition function , usually denoted by the symbol δ . If a finite state machine contains a transition (arrow) from state x to state y , labeled with the input symbol 1 , this means that if the machine is in state x when 1 is read , it goes to state y . We can denote the same action using a transition function, stating that δ ( x , 1) = y . This type of notation is a kind of mathematical shortcut. Putting all this together, we have the formal definition of a finite state machine.

形式定义精确地定义了有限自动机的含义。例如，如果我们回到前面是否允许 0 个接受状态的问题，我们可以看到将F设置为空集Ø给出 0 个接受状态，这是允许的。此外，转换函数δ准确地确定当前状态和输入符号的每种可能组合的下一个状态。这证实了第二个问题的答案，即对于每个可能的输入符号，每个状态必须恰好有一个转换。

The formal definition precisely specifies what we mean by a finite automaton. For example, if we return to the earlier question of whether 0 accepting states are allowed, we can see that setting F to the empty set Ø gives 0 accepting states, which is allowed. Moreover, the transition function δ determines exactly one next state for each possible combination of the current state and the input symbol. This implies a positive answer to the second question, namely that there must be exactly one transition from each state for each possible input symbol.

我们可以通过指定定义 1.5 中列举的五个部分中的每一个来使用形式符号来描述特定的有限自动机。让我们回到前面讨论的有限自动机M ₁，为了方便起见，再次在这里绘制。

We can use formal notation to describe concrete finite automata by specifying each of the five parts enumerated in Definition 1.5. Let us return to the finite automaton M ₁ discussed earlier , redrawn here for convenience.

图 1.6

有限机M ₁

Figure 1.6

Finite state machine M ₁

我们可以通过写 M 1 = ( Q , Σ , δ , q ₁ , F )来正式描述M _{1 ，其中}

We can formally describe M ₁ by writing M1 = ( Q , Σ , δ , q ₁ , F ), where

1. Q = { q ₁ , q ₂ , q ₃ },
2. Q = { q1 , q2 _, q3 } _,​
3. Σ = { 0 , 1 },
4. Σ = { 0 , 1 },
5. δ如表所示

   	0	1
   q 1	q 1	q 2
   q 2	q 3	q 2
   q 3	q 2	q 2

6. δ is described by the table

   	0	1
   q1​	q1​	q2​
   q2​	q3​	q2​
   q3​	q2​	q2​

7. q ₁是初始状态，
8. q ₁ is the initial state,
9. F = { q ₂ }。
10. F = { q ₂ }.

如果A是机器M接受的所有单词的集合，那么我们说A是机器M的语言，并将其写为L ( M )= A。然后我们说M 认识A或M 接受A。因为当我们指代接受单词的机器时，术语“接受”与指语言接受时具有不同的含义，因此在指代语言时，我们更喜欢使用术语“识别”，以避免混淆。

If A is the set of all words accepted by a machine M , then we say that A is the language of M , and we write this as L ( M ) = A . We then say that M recognizes A or that M accepts A . Since the term accepts has different meanings when we refer to the acceptance of words by a machine and different meanings when we refer to the acceptance of languages, we prefer the term recognizes when referring to languages ​​to avoid confusion.

机器可以接受许多单词，但它只能识别一种语言。如果机器不接受任何单词，它仍然会识别一种语言 - 特别是空语言Ø。

The machine can accept many words, but it will always recognize only one language. If the machine does not accept any words, it still recognizes one language – specifically, the empty language Ø .

在我们的例子中，让

In our example, let

A = { w : w至少包含一个 1 和最后一个1之后的偶数个 0 }。

A = { w : w contains at least one 1 and an even number of zeros following the last 1 }.

那么L ( M ₁ ) = A，或者等效地M ₁识别A。

Then L ( M ₁ ) = A , or equivalently M ₁ recognizes A .

有限自动机的例子

Examples of Finite State Machines

例 1.7

Example 1.7

有限机M ₂的状态图如下所示。

The state diagram of the finite state machine M ₂ is shown below .

图 1.8

二状态有限机M _{2的状态图}

Figure 1.8

State diagram for a two-state finite machine M ₂

在形式化描述中，M ₂为({ q ₁ , q ₂ },{ 0 , 1 }, δ , q ₁ ,{ q ₂ })。传递函数δ为

In the formal description M ₂ is ({ q ₁ , q ₂ }, { 0 , 1 }, δ , q ₁ , { q ₂ }). The transition function δ is

	0 0	1 1
q 1 q1​	q 1 q1​	q 2 q2​
q 2 q2​	q 1 q1​	q 2 q2​

重要的是要记住，状态图和M ₂的正式定义包含相同的信息，只是形式不同。如果需要，我们随时可以从一种切换到另一种。

It is important to remember that the state diagram and the formal definition of M ₂ contain the same information, only in a different form. We can always switch from one to the other if necessary.

了解给定机器如何工作的一个好方法是使用一些示例输入进行尝试。当我们进行这样的“实验”来了解机器如何工作时，它的操作方法往往会变得显而易见。对于示例字1101，M ₂机器从初始状态q ₁开始，在读取第一个1后首先移动到状态q ₂，然后依次移动到状态q ₂、q ₁，最后在读取1、0后移动到q ₂，和1。该词被接受，因为q ₂是接受状态。然而，字110使M ₂处于状态q ₁，因此它被拒绝。尝试了其他几个例子后，我们可以看到M _{2接受所有以}1结尾的单词。因此L ( M ₂ ) = { w : w以1结尾}。

A good way to understand how a machine works is to try it out on a few sample inputs. When we do these “experiments” to get a feel for how a machine works, its method of operation often becomes obvious. For example, for word 1101, machine M ₂ starts in state q ₁ and moves first to state q ₂ after reading the first 1 , then to states q ₂ , q ₁ , and finally to q ₂ after reading 1 , 0 , and 1 . The word is accepted because q ₂ is an accepting state. Word 110 , on the other hand , leaves M ₂ in state q ₁ and is therefore rejected. After trying a few more examples, we can see that M ₂ accepts all words ending in 1 . Thus, L ( M ₂ ) = { w : w ends in 1 }.

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例 1.9

Example 1.9

让我们考虑有限机器M ₃。

Consider a finite automaton M ₃ .

图 1.10

二状态有限机M _{3的状态图}

Figure 1.10

State diagram for a two-state finite state machine M ₃

除了接受状态的位置之外，M ₃机器与M ₂类似。像往常一样，机器接受所有单词，当它完成阅读时，它会处于接受状态。注意，由于初始状态也是接受状态，所以M ₃接受空字ε。机器一开始读取空白字，就已经读到最后了；因此，如果初始状态是接受状态，则ε被接受。除了空白之外，本机接受任何以0结尾的单词。因此

Machine M ₃ is similar to M ₂ except for the location of the accepting state. As usual, the machine accepts all words that leave it in the accepting state after it finishes reading them. Note that since the start state is also an accepting state, M ₃ accepts an empty word ε . As soon as the machine starts reading an empty word, it is already at the end; therefore, if the start state is an accepting state, ε is accepted. Apart from an empty word, this machine accepts any word ending in 0 . Thus

L ( M ₃ ) = { w : w是空词ε或以0结尾}。

L ( M ₃ ) = { w : w is the empty word ε or ends with 0 }.

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例 1.11

Example 1.11

下图显示了具有五种状态的M _4机器。

The figure below shows a five-state machine M _{4 .}

图 1.12

有限机M ₄

Figure 1.12

Finite state machine M ₄

M ₄机器包含两个接受状态q ₁和r ₁并在字母表Σ = { a , b } 上进行操作。多项实验表明，它接受单词a、b、aa、bb和bab，但不接受单词ab、ba或bbba。机器从状态s开始，从输入读取第一个符号后，它向左移动到状态q或向右移动到状态r。在这两种情况下，它永远无法返回到初始状态（与前面的示例不同），因为无法从任何其他状态返回到s。如果第一个输入符号是a，则它向左移动并接受以a结尾的单词。同样，如果第一个符号是b，机器向右移动并接受以b结尾的单词。因此M ₄接受所有以a开头和结尾或以b开头和结尾的单词。换句话说，M ₄接受以相同符号开头和结尾的单词。

A machine M ₄ contains two states accepting q ₁ and r ₁ and operates over the alphabet Σ = { a , b }. Several experiments show that it accepts the words a , b , aa , bb , and bab , but not the words ab , ba , or bbba . The machine starts in state s , and after reading the first symbol from the input, it either moves left to states q , or right to states r . In either case, it can never return to the initial state (unlike the previous examples), since there is no way to get from any other state back to s . If the first input symbol is a , it moves left and accepts the word if it ends in a . Similarly, if the first symbol is b , the machine moves right and accepts the word if it ends in b . Thus, M ₄ accepts all words that start and end with a or start and end with b . In other words, M ₄ accepts those words that begin and end with the same symbol.

例 1.13

Example 1.13

图 1.14 显示了在四符号字母表上运行的三状态机M _{5 ，} Σ = { < RESET > , 0 , 1 , 2 }。我们将〈RESET〉视为单个符号。

Figure 1.14 shows a three-state machine M ₅ operating over a four-symbol alphabet, Σ = { 〈 RESET 〉 , 0 , 1 , 2 }. We treat 〈 RESET 〉 as a single symbol.

图 1.14

有限机M ₅

Figure 1.14

Finite state machine M ₅

M ₅机器计算读取的数字符号模3的运行总和。每次遇到符号 <RESET> 时，它都会将计数器重置为 0。如果总和等于 0 模 3，或者在其他情况下，它接受输入单词，当总和是多个数字 3 时。

Machine M ₅ counts the running sum of the numeric symbols read modulo 3. Each time it encounters the symbol 〈 RESET 〉 , it resets the counter to 0. It accepts input if the sum is equal to 0 modulo 3, or in other words, if the sum is a multiple of 3.

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在某些情况下，不可能用状态图来表示有限状态机。当图表太大而无法绘制时，或者如果（如下一个示例中）描述依赖于某些未指定的参数时，可能会发生这种情况。在这种情况下，我们必须依靠正式定义来指定机器。

In some cases it is not possible to represent a finite state machine by a state diagram. This may happen when the diagram is too large to draw, or if (as in the next example) the description depends on some unspecified parameter. In such cases we must rely on a formal definition to specify the machine.

例 1.15

Example 1.15

考虑示例 1.13 的概括。我们仍将使用相同的四符号Σ字母表。令，对于每个i ≥ 1，语言A _i包含所有数字总和为i倍数的单词，并且在求和过程中，只要出现符号<RESET> ，我们就重置计数器。对于每个A _{i ，}我们将呈现一个有限自动机B _i来识别A _i。我们可以将机器B _i正式描述如下：Bi ₌ ( Q _i , Σ , δ _i , q ₀ , { q ₀ })，其中Q _i是i 个状态的集合{ q ₀ , q ₁ , q ₂ , …, q _{i −1} }，我们将设计传递函数δ _{i ，}以便对于每个j，如果B _i处于状态q _j ，则每个q _j的运行和为j模 Let 。

Consider a generalization of Example 1.13. We will still use the same four-symbol alphabet Σ . For each i ≥ 1, let the language A _i contain all words whose sum is a multiple of i , where we reset the counter during the summation whenever the symbol 〈 RESET 〉 appears . For each A _{i ,} we represent a finite automaton B _i that recognizes A _i . The machine B _i can be described formally as follows: B _i = ( Q _i , Σ , δ _i , q ₀ , { q ₀ }), where Q _i is a set of i states { q ₀ , q ₁ , q ₂ , …, q _{i −1} }, and the transition function δ _i is designed so that for each j , if B _i is in state q _j , the running sum is j modulo i. Let for each q _j

δ _i ( q _j , 0 ) = q _j ,

δ _i ( q _j , 0 ) = q _j ,

δ _i ( q _j , 1 ) = q _k，其中k = j + 1 模i，

δ _i ( q _j , 1 ) = q _k , where k = j + 1 modulo i ,

δ _i ( q _j , 2 ) = q _k，其中k = j + 2 模i，

δ _i ( q _j , 2 ) = q _k , where k = j + 2 modulo i ,

δ _i ( q _j , <重置> ) = q ₀。

δ _i ( q _j , 〈 RESET 〉 ) = q ₀ .

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计算的正式定义

Formal definition of computation

到目前为止，我们已经非正式地使用状态图描述了有限状态机，正式地描述为有序五状态机。非正式描述最初更容易理解，但正式定义有助于通过解决非正式描述中可能出现的任何歧义来提供见解的精确性。现在我们将对有限状态机执行的计算（处理）执行相同的操作。我们已经对它如何执行计算有了一个非正式的想法，现在我们将以数学形式将其形式化。

So far, we have described finite state machines informally using state diagrams and with a formal definition as an ordered five. The informal description is easier to initially understand, but the formal definition is useful in ensuring the precision of the observations by resolving any ambiguities that may arise in the informal description. Now we do the same for the computations (processing) performed by the finite state machine. We now have an informal understanding of how it performs its computations, and now we formalize this mathematically.

令M = ( Q , Σ , δ , q ₀ , F ) 为有限自动机，并令w = w ₁ w ₂ …w _n为单词，其中每个w _i是字母表Σ中的符号。如果存在属于Q的状态序列r ₀ , r ₁ , …, r _n，并且满足三个条件，则机器M 接受 w：

Let M = ( Q , Σ , δ , q ₀ , F ) be a finite automaton and let w = w ₁ w ₂ …w _n be a word, where each w _i is a symbol from the alphabet Σ . The machine M accepts w if there is a sequence of states r ₀ , r ₁ , …, r _n belonging to Q , satisfying three conditions:

1. r ₀ = q ₀；
2. r0 = q0 ;_​​
3. δ ( r _i , w _{i +1} ) = r _i +1 对于i = 0, …, n − 1;
4. δ ( r _i , w _{i +1} ) = r _i +1 for i = 0, …, n − 1;
5. r _n ∈ F。
6. r _n ∈ F .

条件 1 表明机器在其初始状态下启动。条件 2 表明机器根据转换函数从一个状态移动到另一个状态。条件 3 表示如果机器以接受状态退出，则机器接受输入。如果A = { w : M接受w }，我们就说M 识别语言 A。

Condition 1 states that the machine starts in the initial state. Condition 2 states that the machine moves from state to state according to a transition function. Condition 3 states that the machine accepts input if it ends in an accepting state. We say that M recognizes a language A if A = { w : M accepts w }.

例 1.17

Example 1.17

让我们假设示例 1.13 中的M ₅机器。让w成为一个词

Let us assume the machine M ₅ from Example 1.13. Let w be a word

10 〈重置〉22 〈重置〉012．

10 〈 RESET 〉22 〈 RESET 〉012 .

那么M ₅按照计算的形式定义接受w ，因为它在处理w 的过程中进入的状态序列是

Then M ₅ accepts w according to the formal definition of computation, since the sequence of states it enters during the processing of w is

q ₀ , q ₁ , q ₁ , q ₀ , q ₂ , q ₁ , q ₀ , q ₀ , q ₁ , q ₀ ,

q ₀ , q ₁ , q ₁ , q ₀ , q ₂ , q ₁ , q ₀ , q ₀ , q ₁ , q ₀ ,

满足所有三个条件。M ₅机器的语言是

that meets all three conditions. The language of the M ₅ machine is

L ( M ₅ ) = { w ：属于w的符号之和为 0 模 3，除了符号<RESET>将计数器重置为 0}。

L ( M ₅ ) = { w : the sum of the symbols belonging to w is 0 modulo 3, except that the symbol 〈 RESET 〉 resets the counter to 0}.

由于M ₅可以识别这种语言，因此它是一种正则语言。

As M ₅ recognizes this language, it is a regular language.

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有限自动机的设计

Designing Finite State Machines

无论是机器还是图形，设计都是一个创造性的过程。因此，它不能简化为简单的配方或配方。尽管如此，可以看出，在设计不同类型的机器时，特定的方法是有帮助的。这包括将我们自己置于我们正在尝试设计的机器的立场上，看看我们应该做什么来完成该机器的任务。假装成一台机器是一种心理技巧，可以帮助我们全身心投入到设计过程中。

Whether it is a machine or a graphic, design is a creative process. As such, it cannot be reduced to a simple recipe or pattern. Nevertheless, it can be seen that a particular approach is helpful in designing different types of machines. This consists in putting ourselves in the place of the machine we are trying to design and seeing how we should proceed in order to accomplish the task of that machine. Pretending to be a machine is a psychological trick that helps to involve our whole mind in the design process.

让我们尝试使用刚刚描述的“阅读器作为自动机”方法来设计一个有限状态机。假设我们有一种特定的语言，并且我们想要设计一个能够识别它的有限自动机。当扮演自动机的角色时，我们收到一个输入单词，我们必须确定它是否是我们的自动机应该识别的语言元素。我们将一次读一个单词符号。在每个符号之后，我们必须判断我们目前看到的单词是否属于该语言。这样做的原因是，就像机器一样，我们不知道单词何时结束，因此我们必须始终准备好答案。

Let us try to design a finite automaton using the "reader as automaton" method just described. Suppose we have a language and we want to design a finite automaton that recognizes it. Playing the role of the automaton, we are given an input word and we need to determine whether it is an element of the language that our automaton is supposed to recognize. We will read the symbols of the word one at a time. After each symbol we need to decide whether the word seen so far belongs to the language. The reason for this is that, like a machine, we do not know when the end of the word will be, so we must always have the answer ready.

首先，为了做出这样的决定，我们需要确定在阅读时我们需要记住我们正在学习的单词的哪些内容。为什么我们不能记住我们所看到的一切？我们必须记住，我们希望表现得像有限状态机，而这种类型的机器只有有限数量的状态，这意味着有限的内存。让我们想象一下，输入数据非常长——比如说从这里到月球——所以我们无法完整地记住它。我们的记忆是有限的——例如一张纸——容量也是有限的。幸运的是，要识别许多语言，我们不必记住整个输入单词。我们只需要记住关键信息即可。到底哪些信息是关键取决于所考虑的特定语言。

To make such decisions, we first need to figure out what we need to remember about the word we are learning as we read it. Why couldn’t we just remember everything we saw? It’s important to remember that we want to behave like a finite-state machine, and such a machine has only a finite number of states, which means a finite memory. Imagine that the input data is extremely long—say, from here to the moon—so we won’t be able to remember all of it. We have a finite memory—a single piece of paper, for example—with a finite capacity. Fortunately, for recognizing many languages, we don’t need to remember the entire input word. We only need to remember the key information. The exact definition of what information is key depends on the particular language being considered.

例如，假设字母表是集合 { 0 , 1 } 并且该语言由包含奇数个 1 的所有单词组成。我们想要构建一个能够识别这种语言的有限机器E ₁ 。假装我们是一个自动机，我们开始逐个符号地读取由 0 和 1 组成的输入字符串。我们是否需要记住到目前为止所见过的整个单词来确定 1 的数量是否为奇数？当然不是。您所需要做的就是记住到目前为止看到的 1 的数量是偶数还是奇数，并在读取新符号时跟踪这些信息。当我们读到1时，我们改变了答案；然而，当我们读取0时，我们将其保持不变。

For example, suppose the alphabet is the set { 0 , 1 } and that the language consists of all words that contain an odd number of ones. We want to construct a finite automaton E ₁ that recognizes this language. Pretending to be the automaton, we begin reading the input sequence of zeros and ones, symbol by symbol. Do we need to memorize the entire word we've seen so far to determine whether the number of ones is odd? Of course not. All we need to do is memorize whether the number of ones we've seen so far is odd or even, and keep track of that information as we read new symbols. When we read a 1 , we change our answer; however, when we read a 0 , we leave it unchanged.

但这如何帮助我们设计E ₁机器呢？一旦我们确定了在阅读过程中需要记住输入单词的哪些信息，我们就可以将该信息呈现为有限的可能性列表。在这种情况下，这些可能性是

But how could this help us design the E ₁ machine ? Once we have determined what information about the input word needs to be remembered while reading it, we can represent this information as a finite list of possibilities. In this case, the possibilities are

1. 即使到目前为止，
2. so far even,
3. 到目前为止很奇怪。
4. so far odd.

然后我们为每种可能性分配一个状态。图 1.18 显示了E ₁的状态。

We then assign a state to each of these possibilities. Figure 1.18 shows the states E ₁ .

图 1.18

两种状态q _par和q _npar

Figure 1.18

Two states q _par and q _npar

然后我们分配转换，检查我们需要如何根据读取的符号从一种可能性转移到另一种可能性。因此，如果状态q _{par r}表示偶数可能性，而q _npar表示奇数可能性，则我们创建导致状态更改为1并保持在原地为0 的转换，如下所示。

_{Next ,} we assign transitions, seeing how we need to go from one possibility to another depending on the symbol read. So if the state q _par represents an even possibility, and q _{npar an odd possibility, then we create transitions that cause a} 1 to change state and a 0 to stay put , as below.

图 1.19

显示能力如何变化的转换

Figure 1.19

Transitions showing how the possibilities change

然后，我们将初始状态设置为与到目前为止读取0 个符号（空字 ε）相关的可能性相对应的状态。在这种情况下，初始状态对应于状态q_对，因为0是偶数。最后，我们将接受状态设置为与我们想要接受输入单词的可能性相对应的状态。我们将q _npar设置为接受状态，因为只要我们看到奇数个单词，我们就想接受一个单词。这些添加如下图所示。

Next, we set the initial state to what should be the state corresponding to the possibility associated with reading 0 symbols  so far (the empty word ε ). In this case, the initial state corresponds to the state q _par , since 0 is an even number. Finally, we set the accepting states to those corresponding to the possibility in which we want to accept the input word. We set q _npar as the accepting state, since we want to accept the word if we see an odd number of ones. These additions are shown in the figure below.

图 1.20

添加初始状态和接受状态

Figure 1.20

Adding Initial and Accepting States

例 1.21

Example 1.21

在这个例子中，我们将展示如何设计一个有限机器E _{2来识别由所有包含}001作为子词的单词组成的正则语言。例如，单词0010、1001、001和11111110011111属于该语言，而单词11和0000则不属于该语言。如果我们是E ₂自动机，我们如何才能尝试识别这种语言？当我们阅读后续符号时，我们最初会跳过所有符号。如果我们遇到0，我们注意到这样一个事实：我们可能刚刚看到了我们正在寻找的模式的三个符号中的第一个，001。如果此时我们看到1，则说明零太少，因此我们返回跳过这些零。然而，如果我们现在看到0，我们应该记住我们刚刚看到了两个模式符号。现在我们需要继续扫描直到看到1。如果找到它，我们就会记住我们找到了该模式并继续读取输入直到结束。

In this example, we will show how to design a finite-state machine E ₂ that recognizes a regular language consisting of all words that contain 001 as a subword. For example, the words 0010 , 1001 , 001 , and 11111110011111 belong to this language, while the words 11 and 0000 do not. How could we try to recognize this language if we were the machine E ₂ ? As we read the symbols, we will initially skip all the ones. If we come across a 0 , we note the fact that we may have just seen the first of three symbols of the pattern 001 . If we see a 1 at this point , there were too few zeros, so we go back to skipping ones. However, if we now see a 0 , we should remember that we have just seen two symbols of the pattern. Now we need to continue scanning until we see a 1 . If we find it, we remember that we found a pattern and continue reading the input until the end.

所以我们有四种可能：

So we have four options:

1. 我们还没有看到任何图案符号，
2. We haven't seen any pattern symbols yet,
3. 我们刚刚看到0 ,
4. We just saw 0 ,
5. 我们刚刚看到00 ,
6. We just saw 00 ,
7. 我们看到了整个001的图案。
8. We saw the entire pattern 001 .

对于这些可能性，我们分配状态q、q ₀、q ₀₀和q ₀₀₁。然后，我们可以通过注意到在状态q中读取1后我们仍处于q状态来分配转换，但读取0会使我们转到q ₀。在状态q _{0 中，}读取1返回到q，但读取0将我们移动到q ₀₀。在状态q _{00 中，}读取1后我们转到q ₀₀₁，但读取0会使我们停留在q ₀₀。最后，在状态q _{001 中，}读取任何符号（0或1）都会导致状态保持在q ₀₀₁中。初始状态为q，唯一接受状态为q ₀₀₁，如图 1.22 所示。

We assign states q , q ₀ , q ₀₀ , and q ₀₀₁ to these possibilities . We can then assign transitions by noting that in state q , reading 1 leaves us in q , but reading 0 leaves us in q ₀ . In state q ₀ , reading 1 leaves us in q , but reading 0 leaves us in q ₀₀ . In state q ₀₀ , reading 1 leaves us in q ₀₀₁ , but reading 0 leaves us in q ₀₀₀ . Finally, in state q _{001 ,} reading any symbol ( 0 or 1 ) leaves us in q ₀₀₁ . The initial state is q , and the only accepting state is q ₀₀₁ , as shown in Figure 1.22.

图 1.22

接受包含001的单词

Figure 1.22

Accepting words containing 001

正则操作

Regular operations

在前两节中，我们介绍并定义了有限自动机和正则语言。我们现在将检查它们的属性。这样做将有助于构建一套方便的技术来设计识别特定语言的机器。该集合还将包括证明某些其他语言非正则（即超出有限自动机的能力）的方法。

In the previous two sections, we introduced and defined finite automata and regular languages. We now turn to the study of their properties. This procedure will help us build a handy set of techniques for designing automata that recognize specific languages. This set will also include methods for proving that certain other languages ​​are not regular (that is, beyond the capabilities of finite automata).

在算术中，基本对象是数字，处理数字的工具是+、×等运算。在计算理论中，这些对象是语言，并且使用专门设计的工具来操作它们。我们将定义三种语言操作，称为正则操作，并使用它们来研究正则语言的属性。

In arithmetic, the basic objects are numbers, and the tools for processing them are operations such as + and ×. In the theory of computation, such objects are languages, and they are operated on using specially designed tools. We will define three operations on languages, called regular operations, and then use them to study the properties of regular languages.

求和运算是已知的。它涉及从语言A和 B中获取所有单词并创建一组。

The operation of sum is already known. It consists of taking all the words from both languages ​​A and  B and creating a single set from them.

串联操作有点困难。它涉及以各种可能的方式将语言A中的单词附加到语言B中的单词的开头，从而创建新语言的单词。

The operation of concatenation is a bit more difficult. It involves attaching words from language A to the beginning of words from B in all possible ways, creating words of a new language.

星号操作（Kleene闭包）与其他两种操作略有不同，因为它适用于一种语言，而不是两种不同的语言。因此，该操作是一个参数，而不是两个参数。它的工作原理是组合语言A中任意数量的单词来创建新语言的单词。由于“任何数字”都包含 0 作为其可能性之一，因此空词ε始终是语言A ^*的元素，无论A是什么。

The asterisk operation (Kleene 's closure ) is slightly different from the other two, because it applies to a single language, not two different ones. This operation is therefore a single-argument operation, not a binary one. It works by combining any number of words from language A to form a word of the new language. Since "any number" includes 0 as one possibility, the empty word ε is always an element of language A ^* no matter what A is .

例 1.24

Example 1.24

令字母表Σ为标准 26 个字母 { a , b , …, z }的集合。如果A = {好,坏} 且B = {男孩,女孩}，则

Let the alphabet Σ be the set of standard 26 letters { a , b , …, z }. If A = { good , bad } and B = { boy , girl }, then

A∪B = {好 ，坏，男孩，女孩} ，

A ∪ B = { good , bad , boy , girl },

A ⚬ B = {好男孩,好女孩,坏男孩,坏女孩},

A ⚬ B = { goodboy , goodgirl , badboy , badgirl },

A ^* = { ε ,好,坏,好好,好坏,坏好,坏坏,好好好,好好坏,好坏好,好坏坏, …}。

A ^* = { ε , good , bad , goodgood , goodbad , badgood , badbad , goodgoodgood , goodgoodbad , goodbadgood , goodbadbad , …}.

■

■

设N = {1, 2, 3, …} 为自然数集。当我们说集合N在乘法下是闭集时，我们的意思是对于任何属于 N 的x和y ，乘积x × y也属于N。相反，N在除法下不是封闭的，因为 1 和 2 属于N，但 1/2 不是。一般来说，如果将某个操作应用于该集合的元素会返回属于该集合的对象，则该对象集合相对于某个操作是封闭的。我们将证明正则语言集对于所有三个正则操作都是封闭的。在第 1.3 节中，我们将展示这些是处理正则语言和理解有限自动机的强大功能的有用工具。我们将从求和运算开始。

Let N = {1, 2, 3, …} be the set of natural numbers. When we say that the set N is closed under multiplication, we mean that for any x and y in N , the product x × y is also in N . By contrast, N is not closed under division, since 1 and 2 are in N , but 1/2 is not. More generally, a set of objects is closed under some operation if applying that operation to the elements of the set returns an object in the set. We will show that the set of regular languages ​​is closed under all three regular operations. In Section 1.3, we will show that these are useful tools for processing regular languages ​​and for understanding the power of finite-state automata. We start with the union operation.

定理1.25

Theorem 1.25

正则语言的类由于并集操作而被封闭。

The class of regular languages ​​is closed under the union operation.

换句话说，如果A ₁和A ₂是正则语言，则A ₁ ∪ A ₂也是正则语言。

In other words, if A ₁ and A ₂ are regular languages, then A ₁ ∪ A ₂ is also a regular language.

证明的概念 我们有正则语言A ₁和A ₂，我们想证明A ₁ ∪ A ₂也是正则语言。由于A ₁和A ₂是正则的，所以我们知道存在某个有限自动机M ₁识别A ₁和某个有限自动机M ₂识别A ₂。为了证明A ₁ ∪ A ₂是正则语言，我们将展示一个识别A ₁ ∪ A ₂的有限自动机M。

Concept of Proof We have regular languages ​​A ₁ and A ₂ and we want to show that A ₁ ∪ A ₂ is also a regular language. Since A ₁ and A ₂ are regular, we know that there is some finite automaton M ₁ that recognizes A ₁ and some finite automaton M ₂ that recognizes A ₂ . To prove that A ₁ ∪ A ₂ is a regular language, we show a finite automaton M that recognizes A ₁ ∪ A ₂ .

这是通过构造证明的。我们将基于M ₁和M _2构造M。为了识别语言的总和，机器M必须接受与M ₁或M ₂接受的完备相同的单词的输入。它通过模拟机器M ₁和M ₂来工作，并在任何一个模拟接受时接受。

This is a proof by construction. We construct M from M ₁ and M ₂ . In order to recognize the union of languages, the automaton M must accept its input for exactly the same words that either M ₁ or M ₂ accepts . It operates by simulating both automatons M ₁ and M ₂ , and accepts whenever either simulation would accept.

如何让机器M模拟M ₁和M ₂？也许它会首先针对给定输入模拟M ₁，然后再模拟M ₂。不过，这里我们必须要小心！一旦输入符号被读取并用于模拟M ₁，我们就无法“倒带”并尝试模拟M ₂。我们需要一种不同的方法。

How can we make an automaton M simulate M ₁ and M ₂ ? Perhaps it will first simulate M ₁ for a given input, and then simulate M ₂ . But we have to be careful here! Once the input symbols have been read and used to simulate M ₁ , we cannot "rewind the tape" and try to simulate M ₂ . We need a different approach.

让我们想象一下我们是一个自动机M。当输入符号一一到达时，我们同时模拟M ₁和M ₂。这样，只需要一次传递输入数据。但是我们如何用有限的内存来跟踪两个模拟呢？我们只需要记住每个自动机在读取输入单词时所处的状态。因此，我们需要记住成对的状态。有多少种可能的状态对？如果M ₁包含k _{1 个}状态并且M ₂有k _{2 个状态，则状态对（一个来自}M ₁另一个来自M ₂ ）的数量是k ₁ × k ₂的乘积。该乘积将是机器M的状态数，每对一个。M中的转换从对到对，更新机器M ₁和M ₂的当前状态。M的接受状态是M ₁或M ₂处于接受状态的那些对。

Imagine that we are an automaton M . As the input symbols arrive one by one, we simulate both M ₁ and M ₂ simultaneously. In this way, only one pass through the input is necessary. But how can we keep track of the two simulations with limited memory? We only need to remember the state that each automaton would be in if it read the input word to that point. Thus, we need to remember the pairs of states. How many possible pairs of states are there? If M ₁ has k ₁ states and M ₂ has k ₂ states , then the number of pairs of states, one from M ₁ and the other from M ₂ , is the product k ₁ × k ₂ . This product will be the number of states of M , one for each pair. Transitions in M ​​go from pair to pair, updating the current state for both M ₁ and M ₂ . The accepting states of M are those pairs in which either M ₁ or M ₂ is in an accepting state.

证明 令M ₁识别A ₁，其中M ₁ = ( Q ₁ , Σ , δ ₁ , q ₁ , F ₁ )，并且M ₂识别A ₂，其中M ₂ = ( Q ₂ , Σ , δ ₂ , q ₂， F ₂）。

Proof Let M ₁ recognize A ₁ , where M ₁ = ( Q ₁ , Σ , δ ₁ , q ₁ , F ₁ ), and M ₂ recognize A ₂ , where M ₂ = ( Q ₂ , Σ , δ ₂ , q ₂ , F ₂ ).

我们构造一个自动机M来识别A ₁ ∪ A ₂，其中M = ( Q , Σ , δ , q ₀ , F )。

We construct an automaton M that recognizes A ₁ ∪ A ₂ , where M = ( Q , Σ , δ , q ₀ , F ).

1. Q = {( r ₁ , r ₂ ) : r ₁ ∈ Q ₁且r ₂ ∈ Q ₂ }。

   该集合是集合Q ₁和Q ₂的笛卡尔积，我们将其写为Q ₁ × Q ₂。它是所有状态对的集合，其中第一个状态来自Q ₁，第二个状态来自Q ₂。

2. Q = {( r ₁ , r ₂ ) : r ₁ ∈ Q ₁ and r ₂ ∈ Q ₂ }.

   This set is the Cartesian product of the sets Q ₁ and Q ₂ and is written as Q ₁ × Q ₂ . This is the set of all pairs of states, the first of which comes from Q ₁ and the second from Q ₂ .

3. Σ ，字母表，与M ₁和M ₂中的相同。在该定理和所有后续类似定理中，为简单起见，我们假设M ₁和M ₂具有相同的输入字母表Σ。如果它们都使用不同的字母Σ ₁和Σ ₂，则该陈述仍然成立。然后我们只需假设Σ = Σ ₁ ∪ Σ ₂来修改证明。
4. Σ , the alphabet, is the same as in M _​​1 and M ₂ . In this theorem and all subsequent similar theorems, we assume for simplicity that both M ₁ and M ₂ have the same input alphabet Σ . The theorem remains true if both used different alphabets, Σ ₁ and Σ ₂ . We would then only have to modify the proof by assuming that Σ = Σ ₁ ∪ Σ ₂ .
5. 我们将传递函数δ定义如下。对于每对 ( r ₁ , r ₂ ) ∈ Q和每个a ∈ Σ，令

   δ (( r ₁ , r ₂ ), a ) = ( δ ₁ ( r ₁ , a ), δ ₂ ( r ₂ , _a ))。

   因此δ接受状态M （本质上是来自M ₁和M ₂的一对状态）以及输入符号，并返回机器M的下一个状态。

6. The transition function δ is defined as follows. For each pair ( r ₁ , r ₂ ) ∈ Q and each a ∈ Σ , let

   δ (( r ₁ , r ₂ ), a ) = ( δ ₁ ( r ₁ , a ), δ ₂ ( r ₂ , _a )).

   So δ takes a state M (which is really a pair of states from M ₁ and M ₂ ) along with an input symbol and returns the next state of the machine M .

7. q ₀是对( q ₁ , q ₂ )。
8. q ₀ is a pair ( q ₁ , q ₂ ).
9. F是其中任何元素是接受M ₁或M ₂的状态的对的集合。

   我们可以把它写成

10. F is the set of pairs in which any of the elements is a state accepting M ₁ or M ₂ .

    We can write this as

F = {( r ₁ , r ₂ ) : r ₁ ∈ F ₁或r ₂ ∈ F ₂ }。

F = {( r ₁ , r ₂ ) : r ₁ ∈ F ₁ or r ₂ ∈ F ₂ }.

该表达式等价于F = ( F ₁ × Q ₂ ) ∪ ( Q ₁ × F ₂ )。 （请注意，这与F = F ₁ × F ₂不同。在这种情况下我们会得到什么？³）

This expression is equivalent to F = ( F ₁ × Q ₂ ) ∪ ( Q ₁ × F ₂ ). (Note that this is not the same as F = F ₁ × F ₂ . What would we get in that case? ³ )

这就完成了识别语言A ₁和A _{2之和的有限自动机}M的构造。这种结构非常简单，因此其正确性显然来自于证明概念中描述的策略。更复杂的结构需要额外的讨论来证明其有效性。此类结构的正确性的正式证明通常是通过归纳法进行的。证明该结构正确性的一个例子可以在定理 1.54 的证明中找到。我们在本书中遇到的大多数结构都相对简单，因此不需要正式的正确性证明。

This completes the construction of the finite automaton M , which recognizes the union of the languages ​​A ₁ and A ₂ . This construction is quite simple, and thus its correctness follows obviously from the strategy described in the proof concept. More complex constructions require additional discussion to show their correctness. A formal proof of correctness for constructions of this type is usually carried out by induction. An example of a proof of the correctness of a construction can be found in the proof of Theorem 1.54. Most of the constructions we will encounter in this book are relatively simple and as such do not require a formal proof of correctness.

我们刚刚证明了两种正则语言的并集是正则语言，因此我们证明了正则语言的类对于并集运算是封闭的。我们现在将处理连接操作，并尝试表明正则语言的类也因该操作而封闭。

We have just shown that the union of two regular languages ​​is a regular language, and thus we have proved that the class of regular languages ​​is closed under the union operation. We now turn to the concatenation operation and try to show that the class of regular languages ​​is also closed under this operation.

定理1.26

Theorem 1.26

正则语言的类由于串联操作而被封闭。

The class of regular languages ​​is closed under the concatenation operation.

换句话说，如果A ₁和A ₂是正则语言，那么A ₁ ⚬ A ₂也是正则语言。

In other words, if A ₁ and A ₂ are regular languages, then A ₁ ⚬ A ₂ is also regular .

为了证明这个定理，我们可以使用类似于求和情况的证明。和之前一样，我们将从有限自动机M ₁和M ₂开始，识别正则语言A ₁和A ₂。然而，这一次，不是构造一个机器M ，当M ₁或M ₂接受输入字时，该机器必须接受输入数据是否可以分为两部分，使得M ₁接受第一个部分，并且M2接受其中的第二个_。问题是M不知道在哪里分割输入数据（即第一部分在哪里结束，第二部分在哪里开始）。为了解决这个问题，我们引入了一种称为非确定性的新技术。

To prove this theorem, we can use something similar to the proof for the sum case. As before, we start with finite automata M ₁ and M ₂ , recognizing regular languages ​​A ₁ and A ₂ . This time, however, instead of constructing an automaton M that accepts its input whenever either M ₁ or M ₂ would accept it, the automaton must accept if the input can be split into two parts such that M ₁ accepts the first and M ₂ accepts the second. The problem is that M does not know where to split the input (that is, where the first part ends and the second begins). To solve this problem, we introduce a new technique called nondeterminism.

1.2

非决定论

1.2

Nondeterminism

非确定性是一个有用的概念，对计算理论产生了很大的影响。到目前为止，在我们的考虑中，每一步的计算都清楚地来自前一步。当机器处于某种状态并读取下一个输入符号时，我们知道它的下一个状态将是什么——它是确定的。我们称之为确定性计算。在非确定性机器中，每个点的下一个状态可以有多种选择。

Nondeterminism is a useful concept that has had a profound impact on the theory of computation. So far, in our discussion, each step of a computation has been a unique consequence of the step before it. When the machine is in a certain state and reads the next input symbol, we know what its next state will be—this is determined. We call this deterministic computation . In a nondeterministic machine, at each point there can be several choices for the next state.

非确定性是确定性的推广，因此每个确定性有限自动机都是非确定性有限自动机。如图 1.27 所示，非确定性有限状态机可以具有其他属性。

Nondeterminism is a generalization of determinism, so every deterministic finite automaton is a nondeterministic finite automaton. As shown in Figure 1.27, nondeterministic finite automatons can have additional properties.

图 1.27

非确定性有限状态机N ₁

Figure 1.27

Nondeterministic finite automaton N ₁

确定性有限自动机（ DFA）和非确定性有限自动机（非确定性有限自动机，NFA）之间的区别乍一看是显而易见的。首先，每个 DFA 状态对于每个字母符号始终只有一个传出转换。图 1.27 所示的 NFA 违反了这一规则。状态q _1对于0有一个传出转换，但对于1有两个传出转换；q ₂有一个0 的传出转换，但没有1 的传出转换。在 NFA 中，对于字母表中的每个符号，状态可以有零个、一个或多个传出转换。

The difference between a deterministic finite automaton (DFA) and a nondeterministic finite automaton (NFA) is obvious at first glance. First, each state of a DFA always has exactly one outgoing transition for each symbol of the alphabet. The NFA shown in Figure 1.27 violates this rule. State q ₁ has one outgoing transition for 0 but two for 1 ; q ₂ has one outgoing transition for 0 but none for 1 . In an NFA, a state can have zero, one, or many outgoing transitions for each symbol of the alphabet.

其次，在 DFA 中，描述转换箭头的标签是字母表中的符号。然而，该 NFA 包含一个标记为ε 的箭头。一般来说，NFA 可能包含由字母元素或ε描述的多个转换。每个状态可以产生零个、一个或多个用标签ε标记的转换（为了简化符号，我们将它们称为ε -转换）。

Second, in a DFA, the labels describing the transition arrows are symbols from the alphabet. However, this NFA contains an arrow labeled ε . In general, an NFA can contain many transitions labeled with elements of the alphabet or ε . Each state can give rise to zero, one, or many transitions labeled ε (for simplicity, we will call them ε -transitions).

NFA 中如何进行处理？假设我们对某个输入词运行 NFA，并到达有几条进一步路径的状态。例如，假设我们处于状态q ₁ NFA N ₁，下一个输入符号是1。读完这个符号后，机器会分裂成许多自身的副本，并并行地追求每种可能性。机器的每个副本都采用可能的路径之一并继续运行。如果出现后来的选择，机器会再次分裂。如果下一个输入符号没有出现在任何离开机器给定副本状态的转换中，则该副本与相应的处理分支一起“死亡”。最后，如果在输入结束时机器的任何剩余副本处于接受状态，则 NFA 接受输入字。

How does processing in NFA work? Suppose we run NFA for some input word and arrive at a state where there are several possible paths. For example, let's say we are in state q ₁ NFA N ₁ and that the next input symbol is 1 . After reading this symbol, the machine splits into multiple copies of itself and pursues each of the possibilities in parallel. Each copy of the machine takes one of the possible paths and continues. If there are later choices, the machine splits again. If the next input symbol does not occur in any of the transitions out of the state of a given copy of the machine, then that copy "dies" along with its corresponding processing branch. Finally, if at the end of the input data any of the remaining copies of the machine is in an accepting state, then the NFA accepts the input word.

如果在传出转换时遇到带有符号ε的状态，则会发生类似的情况。在不读取下一个输入符号的情况下，机器会在每次传出ε转换后分裂成许多副本，并在当前状态中保留一个副本。然后机器像以前一样执行进一步的非确定性处理。

If a state with symbol ε is encountered on an outgoing transition, something similar happens. Without reading another input symbol, the machine splits into multiple copies following each of the outgoing ε -transitions, plus one more remaining in the current state. The machine then performs further nondeterministic processing as before.

非确定性可以被视为一种并行处理，其中可以同时执行多个独立的“进程”或“线程”。当 NFA 拆分以遵循多个选择时，这对应于“分支”为多个子进程，每个子进程单独执行。如果这些进程之一达到接受状态，则所有处理都在接受。

Nondeterminism can be viewed as a type of parallel processing in which multiple independent "processes" or "threads" can execute simultaneously. When an NFA splits to follow multiple choices, this corresponds to "branching" into multiple child processes, each executing separately. If even one of these processes enters an accepting state, the entire processing is accepting.

思考非确定性处理的另一种方法是可能性树的想法。树的根对应于计算的开始。树中的每个分支点对应于机器有多种选择的点。如果至少一个处理分支以接受状态结束，则机器接受，如图 1.28 所示。

Another way to think about nondeterministic computing is in terms of a tree of possibilities. The root of the tree corresponds to the beginning of the computation. Each branching point in the tree corresponds to a point where the machine has several choices. The machine accepts if at least one of the branches of processing ends in an accepting state, as in Figure 1.28.

图 1.28

接受分支的确定性和非确定性处理

Figure 1.28

Deterministic and nondeterministic processing with an accept branch

考虑图 1.27 中所示的NFA N ₁的一些示例运行。输入字010110的N ₁处理如下图所示。

Consider some sample NFA N ₁ runs shown in Figure 1.27. The processing of N ₁ for input word 010110 is shown in the figure below.

对于输入字010110，我们从初始状态q ₁开始并读取第一个符号0。对于0，从状态q ₁只有一个地方可以去- 具体来说，状态q ₁本身- 所以我们留在那里。然后我们读取第二个符号1。在状态q ₁中，符号1有两个选择：保持在q ₁或移动到q ₂。机器以非确定性方式运行，分成两部分并遵循其中的每一个选择。我们可以通过观察机器可能处于的每个状态来跟踪这些可能性。所以我们已经掌握了状态q ₁和q ₂。q ₂状态出现ε跃迁，机器再次分裂；我们将手指放在 q ₂上，并将另一根手指放在q ₃上。所以我们已经掌握了q ₁、q ₂和q ₃。

For input word 010110, we start in the initial state q ₁ and read the first symbol 0 . For 0, there is only one place to go from state q _{1 —namely, state} q ₁ itself —so we stay there. Then we read the second symbol 1 . In state q ₁ , symbol 1 has two choices: staying in q ₁ or going to q ₂ . Operating nondeterministically, the machine splits in two and follows each of these choices. We can keep track of these possibilities by placing fingers on each state the machine could be in. So we have fingers on states q ₁ and q ₂ . From state q ₂ comes an ε -transition and the machine splits again; we keep a finger on state  q ₂ and another on q ₃ . So we have fingers on q ₁ , q ₂ , and q ₃ .

读取第三个符号0后，我们依次检查每个手指。我们将状态q ₁的手指留在原处，将手指从状态q ₂移动到q ₃并拿走处于状态q ₃的手指。对于后者，没有标签0 的转换跟随；这对应于一个简单地“死亡”的进程。此时，我们已经掌握了状态q ₁和q ₃。

After reading the third symbol, 0 , we check each finger in turn. We leave the finger in state q ₁ in place , move the finger in state q ₂ to q ₃ , and take away the one that was in state q ₃ . For the latter, there was no transition with label 0 to follow; this corresponds to a process that simply "died." At this point, we have fingers in states q ₁ and q ₃ .

读完第四个符号1 后，我们将停留在q ₁上的手指分成状态q ₁和q _{2上的手指，然后进一步将}q ₂中的手指分裂为遵循ε转移到q ₃，并且是的手指到目前为止，在q ₃上，我们转向q ₄。我们现在已经掌握了这四个州的情况。

After reading the fourth symbol, 1 splits the finger resting on q ₁ into fingers on states q ₁ and q ₂ , then further splits the finger from q ₂ to follow the ε -transition to q ₃ , and moves the finger that was on q ₃ to q ₄ . We now have fingers on each of the four states.

当我们读取第五个符号1时，处于状态q ₁和q _{3 的}手指导致处于状态q ₁、q ₂、q ₃和q ₄的手指，就像第四个符号的情况一样。状态q ₂的手指被移除。位于q ₄上的手指仍位于q ₄上。现在我们有两个手指放在q ₄上，所以我们可以移除一个手指，因为我们只需要记住q ₄此时是一种可能的状态，而不是由于多种原因它是可能的。

When we read the fifth symbol 1 , the fingers at states q ₁ and q ₃ result in fingers at states q ₁ , q ₂ , q ₃ , and q ₄ , just as in the case of the fourth symbol. The finger at state q ₂ is removed. The finger that was on q ₄ remains on q ₄ . We now have two fingers at q ₄ , so we can remove one, since we only need to remember that q ₄ is a possible state at this point, not that it is possible for multiple reasons.

读取第六个也是最后一个符号0 后，我们将q ₁上的手指留在原处，将状态q ₂移动到q _{3 ，移除}q ₃上的手指，状态q ₄上的手指保持在原位。我们处于单词的末尾，如果其中一根手指处于接受状态，我们可以说机器接受它。手指停留在状态q ₁、q ₃和q ₄上；由于q ₄是接受状态，因此自动机N ₁接受该词。

After reading the sixth and last symbol 0 , we leave the finger on q ₁ in place, move the one from state q ₂ to q ₃ , remove the one that was on q ₃ , and the finger from state q ₄ remains in place. We have reached the end of the word and can say that the automaton accepts it if any of the fingers is in the accepting state. The fingers rest on states q ₁ , q ₃ , and q ₄ ; and since q ₄ is the accepting state, the automaton N ₁ accepts the word.

N ₁对输入010有何作用？我们从q ₁上的手指开始。读取0后，我们仍然只有一根手指放在q ₁上；然而，在1之后，我们将关注状态q ₁、q ₂和q ₃（我们不能忘记ε转变）。在第三个0符号之后，我们将手指从q ₃移开，将手指从q ₂移动到q _{3 ，而}q ₁上的手指保持在原来的位置。此时，我们处于输入字的末尾，并且由于没有手指处于接受状态，因此N ₁丢弃该数据。

What will N ₁ do for input 010 ? We start with a finger on q ₁ . After reading 0 , we still have a finger only on q ₁ ; however, after 1, we will have fingers on states q ₁ , q ₂ , and q ₃ (we cannot forget the ε -transition). After the third symbol 0 , we remove the finger from q ₃ , move the finger from q ₂ to q ₃ , and the finger on q ₁ stays where it is. At this point, we are at the end of the input word, and since none of the fingers have entered an accepting state, N ₁ discards the data.

继续这样实验，我们可以看到N ₁接受所有包含101或11的单词作为子词。

Continuing to experiment in this way, we can see that N ₁ accepts all words that contain 101 or 11 as a subword.

出于多种原因，非确定性有限状态机很有用。正如我们稍后将展示的，任何 NFA 都可以转化为等效的 DFA，并且构建 NFA 有时比直接构建 DFA 更容易。 NFA 可能比其确定性对应物小得多，或者其功能可能更容易理解。有限自动机中的非确定性也是在更强大的计算模型中使用非确定性的一个很好的介绍，因为有限自动机特别容易理解。我们现在将展示 NFA 的一些示例。

Nondeterministic finite automata are useful for many reasons. As we will show later, any NFA can be transformed into an equivalent DFA, and constructing an NFA is sometimes easier than constructing a DFA directly. An NFA can be much smaller than its deterministic counterpart, or its functioning can be easier to understand. Nondeterminism in finite automata is also a good introduction to the use of nondeterminism in stronger models of computation, since finite automata are particularly easy to understand. We will now show some examples of NFAs.

例 1.30

Example 1.30

设A是字母表 { 0 , 1 } 上的一种语言，由所有在倒数第三个位置包含1 的单词组成（例如000100属于A，而0011不属于 A ）。_以下四状态 NFA N2认可A.

Let A be a language over the alphabet { 0 , 1 }, consisting of all words containing 1 in the third last position (i.e., 000100 is a member of A , while 0011 is not). The following four-state NFA N ₂ recognizes A .

图 1.31

NFA N ₂识别语言A

Figure 1.31

NFA N ₂ recognizing language A

您可以想象该 NFA 执行的处理方式是，它保持在状态q ₁，直到它“猜测”它已经是倒数第三个了。只有当输入符号为1时，它才会移动到状态q ₂，然后使用状态q ₃和q ₄来“检查”这个猜测是否正确。

The processing performed by this NFA can be imagined as being in state q ₁ until it “guesses” that it is already in the third position from the end. Only when the input symbol is 1 does it move to state q ₂ , at which point it uses states q ₃ and q ₄ to “check” whether that guess was correct.

正如我们之前提到的，任何 NFA 都可以转换为等效的 DFA，但有时该 DFA 的状态要多很多倍。识别语言A的最小DFA包含八个状态。此外，理解 NFA 的工作原理要容易得多，如下图所示，显示了等效的 DFA。

As we mentioned earlier, any NFA can be converted to an equivalent DFA, but sometimes this DFA has many times more states. The smallest DFA recognizing language A contains eight states. Moreover, understanding the operation of NFA is much easier, as can be seen in the figure below showing an equivalent DFA.

图 1.32

DFA识别语言A

Figure 1.32

DFA recognizing language A

假设我们将ε添加到图 1.31 的N ₂机器中从q ₂到q ₃以及从 q ₃到q _{4 的}转换标签中。因此，两者都将具有标签0、1、ε而不仅仅是0、1。修改后N ₂能识别什么语言？尝试修改图 1.32 中的 DFA 来识别这种语言。

Suppose we have added ε to the labels of the transitions from q ₂ to q ₃ and from  q ₃ to q ₄ in the machine N ₂ of Figure 1.31. Thus, both will have labels 0 , 1 , ε instead of just 0 , 1 . What language will N ₂ recognize after this modification? Try to modify the DFA of Figure 1.32 so that it recognizes this language.

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例 1.33

Example 1.33

下面所示的NFA N ₃具有由单个符号组成的输入字母表 { 0 }。仅包含一个符号的字母表称为一元字母表。

The NFA N ₃ shown below has an input alphabet { 0 } consisting of a single symbol. An alphabet containing only one symbol is called a unary alphabet .

图 1.34

NFA N ₃

Figure 1.34

NFA N ₃

该机器展示了使用ε跃迁的便利性。它接受0 ^k形式的所有单词，其中k是 2 或 3 的倍数。（回想一下，在单词的情况下，上标表示重复，而不是数字的幂）。例如， N ₃接受单词ε、00、000、0000和000000，但不接受0或00000。

This machine demonstrates the convenience of using ε -transitions. It accepts all words of the form 0 ^k , where k is a multiple of 2 or 3. (Recall that in the case of words, the superscript denotes repetition, not exponentiation.) For example, N ₃ accepts the words ε , 00 , 000 , 0000 , and 000000 , but not 0 or 00000 .

你可以想象这台机器是如何工作的，首先猜测是测试 2 的倍数还是 3 的倍数，选择顶部或底部循环，然后检查哪个猜测是正确的。当然，我们可以用没有ε转换或根本没有不确定性的机器来替换这台机器，但是所示的机器是所有识别这种语言的机器中最容易理解的。

The operation of this machine can be imagined as first guessing whether to test for a multiple of 2 or a multiple of 3, choosing either the upper or lower loop, and then checking which guess was correct. Of course, we could replace this machine with one that contains no ε -transitions or any nondeterminism at all, but the machine shown is the easiest to understand of all the automata that recognize this language.

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例 1.35

Example 1.35

NFA 的另一个示例如图 1.36 所示。请随意练习，看看它会接受单词ε、a、baba和baa，但不会接受单词b、bb或babba。稍后我们将使用这台机器来说明将 NFA 转换为 DFA 的过程。

Another example NFA is shown in Figure 1.36. We encourage you to practice on it to see that it will accept the words ε , a , baba , and baa , but will not accept the words b , bb , or babba . We will later use this machine to illustrate the procedure for converting an NFA to a DFA.

图 1.36

NFA N ₄

Figure 1.36

NFA N ₄

非确定性有限自动机的形式定义

Formal definition of a nondeterministic finite automaton

非确定性有限自动机的形式定义类似于前面提出的确定性自动机的定义。两者都有一组状态、一个输入字母表、一个转换函数、一个初始状态和一组接受状态。然而，它们在一个关键方面有所不同。在 DFA 中，转换函数采用一个状态和一个输入符号并返回下一个状态。在 NFA 中，转换函数采用一个状态和一个输入符号或空字，并返回一组可能的下一个状态。为了能够编写正式的定义，我们需要引入额外的符号。对于任何集合Q，我们将使用符号P (Q)作为Q的所有子集的集合。集合P (Q)称为幂集 Q。对于任何字母表Σ，我们将使用符号Σ _ε来表示Σ ∪ { ε }。现在我们可以将 NFA 中传递函数类型的形式形式写为δ : Q × Σ _ε → P (Q)。

The formal definition of a nondeterministic finite state machine is analogous to the definition of a deterministic finite state machine presented earlier. Both have a set of states, an input alphabet, a transition function, an initial state, and a set of accepting states. However, they differ in one key respect. In a DFA, the transition function takes a state and an input symbol and returns the next state. In an NFA, the transition function takes a state and an input symbol or an empty word and returns the set of possible next states . To be able to write the formal definition, we need to introduce some additional notation. For any set Q, we will use the notation P (Q) as the set of all subsets of Q . The set P (Q) is called the power set of Q . For any alphabet Σ, we will use the notation Σ _ε to denote Σ ∪ { ε }. We can now write the formal form of the type of the transition function in an NFA as δ : Q × Σ _ε → P (Q) .

例 1.38

Example 1.38

让我们回忆一下 NFA N ₁：

Let's recall NFA N ₁ :

N ₁的形式化描述为 ( Q , Σ , δ , q ₀ , F )，其中

The formal description of N ₁ is ( Q , Σ , δ , q ₀ , F ), where

1. Q = { q ₁ , q ₂ , q ₃ , q ₄ },
2. Q = { q ₁ , q ₂ , q ₃ , q ₄ },
3. Σ = { 0 , 1 },
4. Σ = { 0 , 1 },
5. δ定义为

   	0	1	Ε
   q 1	{ q 1 }	{ q 1 , q 2 }	关于
   q 2	{ q 3 }	关于	{ q 3 }
   q 3	关于	{ q 4 }	关于
   q 4	{ q 4 }	{ q 4 }	关于

6. δ is defined as

   	0	1	Ε
   q1​	{ q 1 }	{ q1 , q2 }​​	ABOUT
   q2​	{ q 3 }	ABOUT	{ q 3 }
   q3​	ABOUT	{ q 4 }	ABOUT
   q4​	{ q 4 }	{ q 4 }	ABOUT

7. q ₁是初始状态，并且
8. q ₁ is the initial state, and
9. F = { q ₄ }。
10. F = { q ₄ }.

NFA 计算的正式定义与 DFA 类似。令N = ( Q , Σ , δ , q ₀ , F ) 为 NFA ，w为字母表Σ上的一个单词。然后，如果我们可以将w写为w = y ₁ y ₂ …y _m ，则我们说N 接受 w，其中每个y _i是Σ _ε的元素，并且存在一系列状态r ₀ , r ₁ , …, r _m属于Q，满足三个条件：

The formal definition of computation for NFA is analogous to that for DFA. Let N = ( Q , Σ , δ , q ₀ , F ) be an NFA, and w be a word over the alphabet Σ . We then say that N accepts w if we can write w as w = y ₁ y ₂ …y _m , where each y _i is an element of Σ _ε and there is a sequence of states r ₀ , r ₁ , …, r _m belonging to Q , satisfying three conditions:

1. r ₀ = q ₀，
2. r ₀ = q ₀ ,
3. r _{i +1} ∈ δ ( r _i , y _{i +1} ) 对于i = 0, …, m − 1, 并且
4. r _{i +1} ∈ δ ( r _i , y _{i +1} ) for i = 0, …, m − 1, and
5. r _m ∈ F。
6. r _m ∈ F .

条件 1 表明机器在初始状态下启动。条件 2 表明，当机器N处于状态r _i并读取符号y _{i +1时，状态}r _{i +1}是允许的下一个状态之一。请注意，δ ( r _i , y _{i +1} ) 是允许的下一个状态的集合，并且我们要求r _{i +1}是该集合的元素。最后，条件 3 表明如果最后一个状态是接受状态，则机器接受输入字。

Condition 1 states that the machine starts in the initial state. Condition 2 states that state r _{i +1} is one of the legal next states if machine N is in state r _i and reads the symbol y _{i +1} . Note that δ ( r _i , y _{i +1} ) is the set of legal next states, and that we require r _{i +1} to be an element of this set. Finally, condition 3 states that the machine accepts an input word if the last state is an accepting state.

NFA 和 DFA 的等价性

Equivalence of NFA and DFA

确定性和非确定性有限自动机识别同一类语言。这种等价性既令人惊讶又有用。这令人惊讶，因为 NFA 似乎比 DFA 更通用，因此我们期望它们能够识别更多语言。它很有用，因为描述给定语言的 NFA 有时比描述该语言的 DFA 容易得多。

Deterministic and nondeterministic finite automata recognize the same class of languages. This equivalence is both surprising and useful. It is surprising because it might seem that NFAs are more universal than DFAs, so we might expect them to recognize more languages. It is useful because describing an NFA for a given language is sometimes much easier than describing a DFA for that language.

如果两台机器识别相同的语言，我们就说它们是等价的。

We say that two machines are equivalent if they recognize the same language.

定理1.39

Theorem 1.39

对于每个非确定性有限自动机，都有一个等效的确定性有限自动机。

For every nondeterministic finite automaton there is a corresponding equivalent deterministic finite automaton.

证明的概念 如果一种语言被 NFA 识别，我们必须证明存在也识别它的 DFA。这个想法是将 NFA 转换为模拟该 NFA 的等效 DFA。

Proof Concept If a language is recognized by an NFA, we need to show the existence of a DFA that also recognizes it. The idea is to transform the NFA into an equivalent DFA that will simulate this NFA.

让我们回顾一下设计有限自动机的“读者作为自动机”策略。如果我们冒充 DFA，如何模拟 NFA？在处理输入单词时，您需要跟踪什么？在所示的 NFA 示例中，我们通过将手指放在可能在特定点处于活动状态的每个状态来跟踪计算的各个分支。我们根据 NFA 的工作原理通过移动、添加或移除手指来更新模拟。我们所需要跟踪的只是一组“手指”状态。

Recall the reader-as-automaton strategy of finite-state automata design. How could we simulate an NFA if we pretended to be a DFA? What would need to be kept track of as the input word was processed? In the NFA examples shown, we kept track of the various branches of the computation by placing fingers on each state that might be active at a given point. We updated the simulation by moving, adding, or removing fingers in accordance with the way the NFA works. All we had to keep track of was the set of states “with fingers.”

如果k是 NFA 状态的数量，则这些状态有 2 ^k个子集。每个子集对应于 DFA 必须记住的一种可能性，因此模拟 NFA 的 DFA 将有 2k 个^状态。现在我们需要确定正在创建的 DFA 的初始状态和接受状态集以及其转换函数是什么。一旦正式写成，这将更容易讨论。

If k is the number of states of the NFA, then there are 2 ^k subsets of these states for it . Each subset corresponds to one of the possibilities that the DFA must remember, so a DFA simulating the NFA will have 2 ^k states. Now we need to determine the initial state and the set of accepting states of the DFA being created, and what its transition function will be. This will be easier to discuss once we have introduced a formal notation.

证明设N =( Q , Σ , δ , q ₀ , F ) 是一个识别某种语言A 的NFA 。我们将构造一个识别A的DFA M = ( Q' , Σ , δ' , q ₀ ' , F' ) 。在执行完整的构造之前，让我们考虑一个更简单的情况，其中N不包含带有标签ε的转换。稍后我们还将考虑此类ε跃迁。

Proof Let N = ( Q , Σ , δ , q ₀ , F ) be an NFA recognizing some language A . We construct a DFA M = ( Q' , Σ , δ' , q ₀ ' , F' ) recognizing A . Before doing the full construction, let us consider the simpler case in which N does not contain transitions with labels ε . Later we will consider such ε -transitions as well.

1. Q' = P (Q)。

   每个状态M都是N的状态集合。回想一下，P (Q)是Q的所有子集的集合。

2. Q' = P (Q) .

   Each state of M is a set of states of N . Recall that P (Q) is the set of all subsets of Q .

3. 对于R ε Q'和a ε Σ，令δ' ( R , a ) = { q ε Q : q ε δ ( r , a ) 对于某些r ε R }。

   如果R是M 的状态，那么它也是N的状态集合。当M在状态R中读取符号时，这表明属于R的每个状态将被转移到哪里。由于每个状态都可以遍历到可能状态的集合，因此我们对所有这些状态进行并集。该表达式的另一种写法可以是⁴

   

4. For R ∈ Q' and a ∈ Σ let δ' ( R , a ) = { q ∈ Q : q ∈ δ ( r , a ) for some r ∈ R }.

   If R is a state of M , then it is also the set of states of N . When M reads a symbol while in state R , it shows where each state in R will go . Since each state can go to the set of possible states, we perform the union of all these states. Another way to write this expression could be ⁴

5. q ₀ ' = { q ₀ }。

   M开始时的状态对应于仅包含初始状态N的集合。

6. q0 _' = { q0 _} .

   M starts in the state corresponding to the collection containing only the initial state N .

7. F' = { R ∈ Q' : R包含接受N 的状态}。

   如果机器M 此时可能处于的可能状态N之一是接受状态，则机器M接受。

8. F' = { R ∈ Q' : R contains a state accepting N }.

   A machine M accepts if one of the possible states N that it can be in at that moment is an accepting state.

现在我们需要考虑ε跃迁。为此，我们需要一些额外的符号。对于M的任何状态R ，我们将E ( R ) 定义为仅通过遵循ε转移即可从R 的元素到达的状态集合，包括R的元素本身。形式上我们可以这样写：对于R ⊆ Q令E ( R ) = { q : q可以通过以下 0 个或多个ε -转换从R到达q 。

Now we need to include ε -transitions. To do this, we need some additional notation. For any state R of M, we define E ( R ) to be the collection of states that can be reached from elements of R by following only ε -transitions, including elements of R itself . Formally, we can write this: for R ⊆ Q , let E ( R ) = { q : q can be reached from R by following 0 or more ε -transitions}.

现在我们可以修改机器M的转换函数，在每一步之后将额外的手指放置在ε转换可到达的所有状态上。我们通过将δ ( r , a ) 更改为E ( δ ( r , a )) 来得到这一点。意思是

Now we can modify the transition function of the automaton M so as to place additional fingers on all states reachable by ε -transitions after each step. We achieve this by changing δ ( r , a ) to E ( δ ( r , a )). Thus

δ' ( R , a ) = { q ∈ Q : q ∈ E ( δ ( r , a )) 对于某些r ∈ R }。

δ' ( R , a ) = { q ∈ Q : q ∈ E ( δ ( r , a )) for some r ∈ R }.

此外，我们需要修改初始状态M ，以将手指初始放置在从初始状态N沿ε转换可以到达的所有可能状态上。将q ₀ '替换为E ({ q ₀ }) 可提供此效果。因此，我们已经完成了模拟 NFA N的 DFA M的构建。

In addition, we need to modify the initial state M to initially place fingers on all possible states that can be reached from the initial state N along ε -transitions. Replacing q ₀ ' with E ( { q ₀ } ) provides this effect. We have thus completed the construction of the DFA M , which simulates the NFA N .

M设计显然工作正常。在每个输入处理步骤中，M进入与N在该点应处于的状态子集相对应的状态。这样我们的证明就完成了。

The construction of M clearly works correctly. At each input processing step, M enters the state that corresponds to a subset of the states that N should be in at that point. This completes our proof.

定理 1.39 指出任何 NFA 都可以转化为其等价的 DFA。因此，非确定性有限自动机是表征正则语言的另一种方法。我们将这一事实表述为定理 1.39 的结论。

Theorem 1.39 states that every NFA can be transformed into an equivalent DFA. Thus, nondeterministic finite automata are an alternative way of characterizing regular languages. We will formulate this fact as a corollary of Theorem 1.39.

应用1.40

Motion 1.40

当且仅当一种语言被某种非确定性有限自动机识别时，它才是正则语言。

A language is regular if and only if it is recognized by some nondeterministic finite automaton.

“当且仅当”条件的一个方向表明，如果某种语言被某些 NFA 识别，则该语言是正则语言。定理 1.39 表明任何 NFA 都可以转换为等效的 DFA。因此，如果 NFA 识别某种语言，某种 DFA 也会识别，因此该语言是正则的。 “当且仅当”条件的第二个方向认为，只要某种语言被某个 NFA 识别，它就是正则语言。换句话说，如果一种语言是正则语言，那么就有一些 NFA 可以识别它。这个条件显然是满足的，因为每个正则语言都有一个可以识别它的 DFA，并且每个 DFA 也是一个 NFA。

One direction of the if-and-only-if condition states that a language is regular if it is recognized by some NFA. Theorem 1.39 shows that every NFA can be converted into an equivalent DFA. So if an NFA recognizes a language, then so does some DFA, and hence that language is regular. The other direction of the if-and-only-if condition states that a language is regular if it is recognized by some NFA. In other words, if a language is regular, then there is some NFA that recognizes it. This condition is obviously satisfied, since every regular language has a DFA that recognizes it, and every DFA is also an NFA.

例 1.41

Example 1.41

让我们尝试使用例 1.35 中所示的N ₄机器来说明定理 1.39 的证明中所示的 NFA 到 DFA 的转换过程。为了清楚起见，我们将机器N ₄的状态标签更改为 {1, 2, 3}。因此，在形式化表示中，N ₄ = ( Q , {a, b }, δ , 1, {1})，其中状态集合Q是集合 {1, 2, 3}，如图 1.42 所示。

Let us illustrate the procedure for converting NFA to DFA, given in the proof of Theorem 1.39, using the N ₄ machine shown in Example 1.35. For clarity, we have changed the labels of the states of N ₄ to {1, 2, 3}. Thus, in formal notation, N ₄ = ( Q , {a, b }, δ , 1, {1}), with the set of states of Q being {1, 2, 3}, as shown in Figure 1.42.

为了构造一个等价于N 4的DFA D _{，我们必须首先定义}D的状态。 N ₄具有三个状态 {1, 2, 3}，因此我们将构造具有八个状态的D，每个状态对应N ₄状态的每个子​​集。我们将描述每个状态D及其相应的子集。因此，状态集D为

To construct a DFA D that is equivalent to N ₄ , we first need to determine the states of D . N ₄ has three states, {1, 2, 3}, so we will construct D with eight states, one for each subset of the states of N ₄ . We will describe each state of D by its corresponding subset. Thus, the set of states of D is

{ Ø , {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}}。

{ Ø , {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}}.

图 1.42

NFA N ₄

Figure 1.42

NFA N ₄

然后我们确定初始状态和接受状态D。初始状态是集合E ({1})，即从状态1通过ε-转移和状态1本身可到达的状态集合； ε跃迁从 1 引导至 3，因此E ({1}) = {1, 3}。新的接受状态是包含N ₄台机器的接受状态的子集。因此，它是集合{{1}, {1,2}, {1,3}, {1,2,3}}。

Next, we establish the initial state and the accepting states D . The initial state is the set E ({1}), i.e. the set of states reachable from state 1 via ε -transitions and state 1 itself; the ε -transition leads from 1 to 3, so E ({1}) = {1, 3}. The new accepting states are those subsets that contain an accepting state of the machine N ₄ . This is therefore the set {{1}, {1,2}, {1,3}, {1,2,3}}.

最后，我们需要确定D的传递函数。每个状态D都会导致一个输入a的位置和一个输入b的位置。我们将使用几个例子来说明D中安排转换的过程。

Finally, we need to define a transition function for D . Each state of D leads to one location for input a and one for input b . We will illustrate the process of arranging transitions in D with several examples.

在机器D 中，状态 {2} 导致输入a的 {2,3} ，因为在N ₄中，状态 2 导致输入a的状态 2 和 3 ，并且我们无法从状态 2 和 3 进一步沿着ε转换。对于输入b，状态 {2} 会导致状态 {3} ，因为在N _{4中，对于输入}b，状态 2 只能转换到状态 3 ，并且我们无法沿着ε转换从状态 3 进一步前进。

In automaton D, state {2} leads to state {2,3} for input a , because in N ₄ state 2 leads to both states 2 and 3 for input a and we cannot go any further from states 2 and 3 along ε -transitions. State {2} leads to state {3} for input b , because in N ₄ state 2 has a transition only to state 3 for input b and we cannot go anywhere further from state 3 along ε -transitions.

输入a 的状态 {1}导致Ø，因为N ₄中的状态 1没有带有标签a 的传出转换。反过来，对于b，它会导致状态 {2}。请注意，定理 1.39 中描述的过程指定我们在读取每个输入符号后遵循ε转换。基于在读取每个输入符号之前遵循ε转换原则的替代过程也同样有效，但我们不会在本示例中对其进行说明。

State {1} for input a leads to Ø , since state 1 in N ₄ has no outgoing transitions with label a . On the other hand, for b it leads to state {2}. Note that the procedure described in Theorem 1.39 specifies that we follow the ε -transitions after reading each input symbol. An alternative procedure based on following the ε -transitions before reading each input symbol would work equally well, but we will not illustrate it in this example.

对于a ，状态 {3} 会导致状态 {1,3} ，因为在N _{4中，对于}a，状态 3 会过渡到 1 ，并且可以通过ε过渡从 1 过渡到 3。b的状态 {3}导致Ø。

State {3} leads to state {1,3} for a , since in N ₄ state 3 has a transition to 1 for a , and from 1 it is possible to transition to 3 via an ε -transition. State {3} for b leads to Ø .

a的状态 {1,2}转换为 {2,3}，因为 1 对于该输入没有转换，2 指向a的状态 2 和 3 ，并且它们都没有进一步的ε转换。对于b，状态 {1,2} 导致 {2,3}。继续这样，我们得到如图1.43所示的机器D的图。

The state {1,2} for a has a transition to {2,3}, since 1 has no transitions for this input, 2 indicates for a both states 2 and 3 and neither of them has any further ε -transitions. For b the state {1,2} leads to {2,3}. Continuing in this way, we obtain the diagram of the automaton D shown in Figure 1.43.

图 1.43

DFA D相当于NFA N ₄

Figure 1.43

DFA D equivalent to NFA N ₄

我们可以通过注意到没有任何转换导致状态 {1} 和 {1, 2} 来简化该机器，因此可以在不影响机器效率的情况下删除它们。这样我们就得到了下面的图。

We can simplify this machine by noting that no transitions lead to states {1} and {1, 2}, so they can be removed without affecting the efficiency of the machine. This gives us the following figure.

图 1.44去除不必要状态后的

DFA D

Figure 1.44

DFA D after removing unnecessary states

暂停正常运营

Closure due to regular operations

我们现在可以回到我们在 1.1 节中开始讨论的由于正则操作而关闭正则语言类的问题。我们的目标是证明正则语言的并集、串联和 Kleene 闭包仍然是正则的。当发现串联操作太复杂而无法处理时，我们放弃了最初的尝试。使用非决定论将使证明变得更加容易。

We can now return to the problem of closure of the class of regular languages ​​under regular operations, which we started in Section 1.1. Our goal is to prove that the union, concatenation, and Kleene closure of regular languages ​​are still regular. We abandoned the original attempt at this task when it turned out that dealing with the concatenation operation was too complicated. Using nondeterminism will make the proof much easier.

首先，我们再次考虑为了总数而关闭。我们之前通过构造笛卡尔积来确定性地同时模拟两台机器，从而证明了并集的封闭性。我们现在提出一个新的证明来说明基于非确定性的技术。重新审视第 45 页上显示的第一个证明可能很有价值，因为它将显示新证明是多么容易和直观。

First, let us reconsider closure under union. We proved closure under union earlier by simulating both machines simultaneously deterministically by constructing the Cartesian product. We now present a new proof to illustrate the technique based on nondeterminism. It may be valuable to review the first proof shown on page 45, to see how much easier and more intuitive the new proof is.

定理1.45

Theorem 1.45

正则语言的类由于并集操作而被封闭。

The class of regular languages ​​is closed under the union operation.

证明的概念我们有正则语言A ₁和A ₂，我们想证明A ₁ ∪ A ₂是正则语言。这个想法是为语言A ₁和A ₂采用两个 NFA N ₁和N ₂，并将它们组合成一个新的 NFA N。

Concept of the Proof We have regular languages ​​A ₁ and A ₂ and we want to prove that A ₁ ∪ A ₂ is regular. The idea is to take two NFAs, N ₁ and N ₂ for the languages ​​A ₁ and A ₂ , and combine them into one new NFA N .

如果N ₁或N ₂接受它，机器N必须接受它的单词。新机器有一个新的初始状态，它通过ε转换分支到旧机器的初始状态。通过这种方式，新机器不确定地猜测两台机器中的哪一台应该接受输入。如果其中一台接受输入，机器N也会接受它。

Machine N must accept its word if N ₁ or N ₂ accepts it. The new machine has a new initial state that branches to the initial states of the old machines via ε -transitions. In this way, the new machine nondeterministically guesses which of the two machines is to accept the input it receives. If one of them accepts the input, machine N also accepts it.

我们在下图中展示了这个结构。在左边，我们用较大的圆圈标记了机器N ₁和N ₂的启动和接受状态，并用较小的圆圈标记了几个附加状态。右侧我们展示了如何通过包含额外的转换将N ₁和N ₂机器组合成N。

We have shown this construction in the figure below. On the left, we have marked the start and accepting states of machines N ₁ and N ₂ with larger circles, and several additional states with smaller ones. On the right, we show how to combine machines N ₁ and N ₂ into N by adding additional transitions.

证据

Evidence

令 N ₁ = ( Q ₁ , Σ , δ ₁ , q ₁ , F ₁ ) 识别语言A ₁且

Let N ₁ = ( Q ₁ , Σ , δ ₁ , q ₁ , F ₁ ) recognize the language A ₁ and

N ₂ = ( Q ₂ , Σ , δ ₂ , q ₂ , F ₂ ) 识别语言A ₂。

N ₂ = ( Q ₂ , Σ , δ ₂ , q ₂ , F ₂ ) recognizes the language A ₂ .

我们构造N = ( Q , Σ , δ , q ₀ , F ) 来识别A ₁ ∪ A ₂。

We construct N = ( Q , Σ , δ , q ₀ , F ) which is supposed to recognize A ₁ ∪ A ₂ .

1. Q = { q ₀ } ∪ Q ₁ ∪ Q ₂。

   机器N的状态是所有状态N ₁和N ₂，并补充有新的初始状态q ₀。

2. Q = { q ₀ } ∪ Q ₁ ∪ Q ₂ .

   The states of the automaton N are all the states of N ₁ and N ₂ , supplemented with a new initial state q ₀ .

3. 状态q _0是N的初始状态。
4. State q ₀ is the initial state of N .
5. 接受F = F ₁ ∪ F ₂的状态集。

   接受N的状态是所有接受N ₁和N _{2 的}状态。因此，如果输入词被任何自动机N ₁或N _{2接受，则}N接受该输入词。

6. The set of accepting states F = F ₁ ∪ F ₂ .

   The accepting states of N are all accepting states of N ₁ and N ₂ . Thus N accepts an input word if it is accepted by either of the automata N ₁ or N ₂ .

7. 我们定义δ，以便对于任何q ∈ Q和任何a ∈ Σ _ε，

   

8. We define δ so that for any q ∈ Q and any a ∈ Σ _ε ,

我们现在可以证明由于串联而导致的闭包。回想一下，以前，如果不使用非确定性，完成证明就太困难了。

We can now prove closure under concatenation. Recall that earlier, without using nondeterminism, it turned out to be too difficult to complete the proof.

定理1.47

Theorem 1.47

正则语言的类由于串联操作而被封闭。

The class of regular languages ​​is closed under the concatenation operation.

证明的概念给定正则语言A ₁和A ₂，我们想证明A ₁   ⚬  A ₂也是正则语言。这个想法是为语言A ₁和A ₂使用两个 NFA - N ₁和N ₂ ，并将它们组合成一个新的 NFA N，类似于并集，但这次组合将以不同的方式完成，如图1.48所示。

Concept of the Proof We are given regular languages ​​A ₁ and A ₂ and we want to prove that A ₁   ⚬  A ₂ is also regular. The idea is to take two NFAs – N ₁ and N ₂ – for the languages ​​A ₁ and A ₂ and combine them into a new NFA N , similarly to the union case, but this time the combination happens in a different way, as shown in Figure 1.48.

我们选择初始状态N ₁作为机器N的初始状态。接受N _{1 的}状态接收额外的ε -转换，每当N ₁处于接受状态时，这允许非皮肤分支到N _{2 ，这意味着我们已经找到了从语言}A ₁创建单词的初始输入片段。接受N台机器的状态是接受N ₂台机器的状态。通过这种方式，该机器接受输入可以分为两部分的时间，第一部分由N ₁接受，第二部分由N ₂接受。我们可以将N视为一台能够非机器性地猜测在哪里划分的机器。

As the initial state of machine N we choose the initial state N ₁ . The accepting states of N ₁ receive additional ε -transitions that allow nonderministic branches to N ₂ whenever N ₁ is in an accepting state, meaning that we have found the initial fragment of input that forms a word of language A ₁ . The accepting states of machine N are the accepting states of machine N ₂ . Thus, this machine accepts whenever the input can be split into two parts, with the first being accepted by N ₁ and the second by N ₂ . We can think of N as a machine that nonderministically guesses where the split should be made.

图 1.48识别A ₁  A _2的N

的构造  

Figure 1.48

Construction of N recognizing A ₁  A ₂  

证据

Evidence

令 N ₁ = ( Q ₁ , Σ , δ ₁ , q ₁ , F ₁ ) 识别语言A ₁且

Let N ₁ = ( Q ₁ , Σ , δ ₁ , q ₁ , F ₁ ) recognize the language A ₁ and

N ₂ = ( Q ₂ , Σ , δ ₂ , q ₂ , F ₂ ) 识别语言A ₂。

N ₂ = ( Q ₂ , Σ , δ ₂ , q ₂ , F ₂ ) recognizes the language A ₂ .

我们构造N = ( Q , Σ , δ , q ₁ , F ₂ ) 来识别A ₁   ⚬  A ₂。

We construct N = ( Q , Σ , δ , q ₁ , F ₂ ) which is supposed to recognize A ₁   ⚬  A ₂ .

1. Q = Q ₁ ∪ Q ₂。

   机器N的状态是机器N ₁和N ₂的所有状态。

2. Q = Q ₁ ∪ Q ₂ .

   The states of the automaton N are all the states of the automatons N ₁ and N ₂ .

3. 状态q _{1与初始状态}N ₁相同。
4. The state q ₁ is the same as the initial state N ₁ .
5. 接受F ₂的状态与接受N _{2 的}状态相同。
6. The states accepting F ₂ are the same as the states accepting N ₂ .
7. 我们定义δ，以便对于每个q ∈ Q和每个a ∈ Σ _ε，
8. We define δ so that for every q ∈ Q and for every a ∈ Σ _ε ,

定理1.49

Theorem 1.49

正则语言的类由于星号运算符而被封闭（Kleene 闭包）。

The class of regular languages ​​is closed under the asterisk operator (Kleene closure ) .

证明的概念给定一个正则语言A ₁，我们想证明A 1 ^*也是正则语言。我们将使用识别语言A _{1的 NFA} N ₁并将其修改为识别语言A 1 ^*，如下图所示。生成的 NFA N将接受输入单词，只要它可以分成多个部分，使得每个部分都被N ₁接受。

Concept of Proof We are given a regular language A ₁ and we want to prove that A 1 ^* is also regular. We will use an NFA N ₁ that recognizes language A ₁ and modify it to recognize language A 1 ^* , as in the figure below. The resulting NFA N will accept an input word if it can be split into multiple parts such that each part is accepted by N ₁ .

我们将构建一个N自动机作为 N 1自动机的修改，其中我们将添加从接受状态到初始状态的ε转换。这样，当处理到达N ₁接受的片段的末尾时，N机器将有机会跳回初始状态并尝试读取N ₁接受的另一个片段。此外，我们需要修改N以接受ε，它始终是语言A 1 ^*的元素。一个可能的（不是最好的）想法是简单地将初始状态添加到接受状态集中。这种方法无疑将ε添加到识别的语言中，但也可能添加其他不需要的单词。练习 1.15 给出了这个概念的缺陷的一个例子。解决这个问题的方法是添加一个新的初始状态，它也是接受状态并包含到旧初始状态的ε转换。该解决方案实现了向语言添加ε的预期效果，但没有添加任何超出它的内容。

We will construct the automaton N as a modification of the automaton N 1 , in which we add ε -transitions leading from the accepting states to the initial state. In this way, when processing reaches the end of a fragment accepted by N ₁ , the automaton N will have the option of jumping back to the initial state and trying to read another fragment accepted by N ₁ . In addition, we must modify N so that it accepts ε , which is always an element of the language A 1 ^* . One possible (not very good) idea is to simply add the initial state to the set of accepting states. This approach undoubtedly adds ε to the recognized language, but it may also add other, undesirable words. Exercise 1.15 illustrates the flaw in this idea with an example. A way to solve this problem is to add a new initial state that is also an accepting state and that contains an ε -transition to the old initial state. This solution achieves the desired effect of adding ε to the language, but without adding anything else to it.

图 1.50识别语言A的N

的设计^*

Figure 1.50

Construction of N recognizing language A ^*

证明设N ₁ = ( Q ₁ , Σ , δ ₁ , q ₁ , F ₁ ) 识别语言A ₁。

Proof Let N ₁ = ( Q ₁ , Σ , δ ₁ , q ₁ , F ₁ ) recognize the language A ₁ .

我们构造N = ( Q , Σ , δ , q ₀ , F ) 来识别语言A 1 ^*。

We construct N = ( Q , Σ , δ , q ₀ , F ) recognizing the language A 1 ^* .

1. Q = { q ₀ } ∪ Q ₁。

   N台机器的状态是N ₁的状态加上一个新的初始状态。

2. Q = { q ₀ } ∪ Q ₁ .

   The states of the automaton N are the states of N ₁ plus the new initial state.

3. 状态q ₀是新的初始状态。
4. State q ₀ is the new initial state.
5. F = { q ₀ } ∪ F ₁。

   接受状态是旧的一组接受状态加上新的起始状态。

6. F = { q ₀ } ∪ F ₁ .

   The accepting states are the old set of accepting states plus the new initial state.

7. 我们定义δ，以便对于任何q ∈ Q和任何a ∈ Σ _ε，

   

8. We define δ so that for any q ∈ Q and any a ∈ Σ _ε ,

1.3

正则表达式

1.3

Regular expressions

在算术中，我们可以使用 + 和 × 运算来构建表达式，例如

In arithmetic, we can use the + and × operations to build expressions like

(5 + 3) × 4。

(5 + 3) × 4 .

类似地，我们可以使用正则运算来构建描述语言的表达式；我们称它们为正则表达式。一个例子是：

Similarly, we can use regular operations to build expressions that describe languages; we call these regular expressions . An example would be:

( 0 ∪ 1 ) 0 ^*。

( 0 ∪ 1 ) 0 ^* .

显示的算术表达式的值为数字 32。正则表达式的值为 language。在这种情况下，该值是一种语言，由所有以0或1开头，后跟任意数量的零的单词组成。我们可以通过将这个表达式分成几部分来得到这个结果。首先，符号0和1是集合{0}和{1}的缩写。因此，( 0 ∪ 1 ) 意味着({ 0 } ∪ { 1 })。这部分的值为语言{ 0,1 }。0 ^*部分表示 { 0 } ^*，其值是由包含任意数量的零的所有单词组成的语言。其次，就像代数中的 × 符号一样，串联符号 ⚬通常隐含在正则表达式中。所以⁽ 0∪1 ) 0 ^*实际上 是( 0∪1 ) ⚬0 *的缩写。将两个部分的单词 join 连接起来以创建整个表达式的值。

The value of the arithmetic expression shown is the number 32. The value of the regular expression is the language. In this case, the value is the language of all words starting with 0 or 1 , followed by any number of zeros. We can obtain this result by splitting the expression into parts. First, the symbols 0 and 1 are shorthand for the sets { 0 } and { 1 }. Thus, ( 0 ∪ 1 ) means ({ 0 } ∪ { 1 }). The value of this part is the language { 0 , 1 }. The part 0 ^* means { 0 } ^* , and its value is the language of all words containing any number of zeros. Second, like the symbol × in algebra, the concatenation symbol ⚬ is often implicit in regular expressions. So ( 0 ∪ 1 ) 0 ^* is essentially a shorthand for ( 0 ∪ 1 ) ⚬ 0 ^* . Concatenation joins the words from both parts to form the value of the entire expression.

正则表达式在计算应用中发挥着重要作用。在基于文本的应用程序中，用户可能想要搜索与某些模式匹配的字符串。正则表达式提供了描述此类模式的有效方法。 UNIX 系统上的awk和grep等实用程序、Perl 等现代编程语言以及文本编辑器都提供了使用正则表达式描述模式的机制。

Regular expressions play an important role in computer science. In text-based applications, users may wish to search for strings that match specific patterns. Regular expressions provide an efficient method for describing such patterns. Utilities such as awk and grep on UNIX systems, modern programming languages ​​such as Perl, and text editors all provide mechanisms for describing patterns using regular expressions.

例 1.51

Example 1.51

正则表达式的另一个例子是

Another example of a regular expression is

( 0 ∪ 1 ) ^*。

( 0 ∪ 1 ) ^* .

我们从语言( 0∪1 )开始并使用^*操作。该表达式的值是由所有可能的零和一组成的字符串组成的语言。如果Σ = { 0 , 1 }，我们可以使用Σ作为正则表达式 ( 0 ∪ 1 )的简写。更一般地，如果Σ是任意字母表，则正则表达式Σ描述该字母表之上的所有长度为1的单词的语言，并且Σ ^*描述该字母表之上的所有（任意长度）单词的语言。类似地，Σ ^* 1是一种包含所有以1结尾的单词的语言。语言 ( 0 Σ ^* ) ∪ ( Σ ^* 1 ) 由所有以0开头或以1结尾的单词组成。

Here we start with the language ( 0 ∪ 1 ) and apply the ^* operation . The value of this expression is the language of all possible strings consisting of zeros and ones. If Σ = { 0 , 1 }, we can use Σ as a shorthand for the regular expression ( 0 ∪ 1 ). More generally, if Σ is an arbitrary alphabet, the regular expression Σ describes the language of all words of length 1 over this alphabet, and Σ ^* describes the language of all words (of arbitrary length) over this alphabet. Similarly, Σ ^* 1 is the language of all words ending in 1 . The language ( 0 Σ ^* ) ∪ ( Σ ^* 1 ) consists of all words starting with 0 or ending with 1 .

在算术中，我们说 × 优先于 + 来指定如果有选择，我们首先执行 × 操作。因此，在表达式2+3×4中，3×4项在加法之前执行。要先进行加法，我们需要添加括号，得到表达式 (2 + 3) × 4。在正则表达式中，先进行星号运算，然后进行连接，最后进行求和，除非使用括号改变了顺序。

In arithmetic, we say that × takes precedence over + to indicate that if there is a choice, we perform the × operation first. Thus, in the expression 2 + 3 × 4 , the 3 × 4 term is performed before the addition. To perform the addition first, we add the parentheses, obtaining the expression (2 + 3) × 4 . In regular expressions, the asterisk operation is performed first, then the concatenation, and finally the sum, unless the order is changed by using parentheses.

正则表达式的正式定义

Formal definition of a regular expression

正则表达式ε和Ø不应混淆。表达式ε表示一种包含一个单词的语言（具体来说，是一个空单词），而Ø表示一种不包含单词的语言（空语言）。

The regular expressions ε and Ø should not be confused . The expression ε represents a language containing one word—specifically, the empty word—whereas Ø represents a language containing no words at all (an empty language).

显然，使用正则表达式本身来定义正则表达式的概念存在危险。如果是这样的话，我们在定义时就会陷入恶性循环，这样的定义将是不正确的（逻辑错误）。然而，R ₁和R ₂总是小于R。因此，我们实际上是通过更小的正则表达式来定义正则表达式，从而避免循环。这种定义称为归纳定义。

Apparently, we have here the danger of defining the concept of a regular expression in terms of itself. If this were the case, we would have a vicious circle in defining , and such a definition would be incorrect (a logical fallacy). However, R ₁ and R ₂ are always smaller than R . Thus, in fact, we define regular expressions by smaller regular expressions, and thus avoid cyclicity. A definition of this kind is called an inductive definition .

表达式中的括号可以省略。如果是这种情况，则按照优先顺序进行处理：星号（Kleene 闭包）、串联，最后并集。

Parentheses in expressions may be omitted. If so, processing is in order of precedence: asterisk (Kleene 's closure ), concatenation, and finally union.

为了方便起见，我们将使用符号R ⁺作为RR ^*的缩写。换句话说，虽然R ^*包含R中 0 个或多个单词串联的所有单词，但语言R ^+包含R中1 个或多个单词串联的所有单词。所以R ⁺ ∪ ε = R ^*。此外，我们将R ^k定义为将k 个R表达式相互连接的快捷方式。

For convenience, we will use the notation R ⁺ as shorthand for RR ^* . In other words, if R ^* contains all words that are the concatenation of 0 or more words from R , then the language R ⁺ contains all words that are the concatenation of 1 or more words from R . So R ⁺ ∪ ε = R ^* . We further define R ^k as shorthand for the concatenation of k expressions of R with themselves.

当我们需要区分R正则表达式和它所描述的语言时，我们将使用符号L ( R ) 来表示R语言。

Whenever we need to distinguish between a regular expression R and the language it describes, we will use the notation L ( R ) to denote the language R .

实施例 1.53

Example 1.53

在以下几点中，我们假设字母表Σ为 { 0 , 1 }。

In the following steps we assume that the alphabet Σ is { 0 , 1 }.

1. 0 ^* 10 ^* = { w : w包含单个1 }。
2. 0 ^* 10 ^* = { w : w contains a single 1 }.
3. Σ ^* 1 Σ ^* = { w : w至少包含一个1 }。
4. Σ ^* 1 Σ ^* = { w : w contains at least one 1 }.
5. Σ ^* 001 Σ ^* = { w : w包含子字001 }。
6. Σ ^* 001 Σ ^* = { w : w contains substring 001 }.
7. 1 ^* ( 01 ⁺ ) ^* = { w : w 中的每个0后面至少有一个1 }。
8. 1 ^* ( 01 ⁺ ) ^* = { w : each 0 in word w is followed by at least one 1 }.
9. ( ΣΣ ) ^* ={ w : w是偶数长度的字} ⁵．
10. ( ΣΣ ) ^* = { w : w is an even-length word } ⁵ .
11. ( ΣΣΣ ) ^* = { w : w的长度是 3 的倍数}。
12. ( ΣΣΣ ) ^* = { w : length of w is a multiple of 3}.
13. 01 ∪ 10 = { 01 , 10 }。
14. 01 ∪ 10 = { 01 , 10 }.
15. 0 Σ ^* 0 ∪ 1 Σ ^* 1 ∪ 0 ∪ 1 = { w : w以相同符号开始和结束}。
16. 0 Σ ^* 0 ∪ 1 Σ ^* 1 ∪ 0 ∪ 1 = { w : w starts and ends with the same symbol}.
17. ( 0 ∪ ε ) 1 ^* = 01 ^* ∪ 1 ^*。

    表达式0 ∪ ε描述了语言 { 0 , ε }，因此串联操作会在1 ^*的每个单词之前添加0或ε。

18. ( 0 ∪ ε ) 1 ^* = 01 ^* ∪ 1 ^* .

    The expression 0 ∪ ε describes the language { 0 , ε }, so the concatenation operation adds either 0 or ε before each word of 1 ^* .

19. ( 0 ∪ ε )( 1 ∪ ε ) = { ε , 0 , 1 , 01 }。
20. ( 0 ∪ ε )( 1 ∪ ε ) = { ε , 0 , 1 , 01 }.
21. 1 ^* Ø = Ø。

    空集与任何集的串联总是产生空集。

22. 1 ^* Ø = Ø .

    The concatenation of the empty set with any set always produces the empty set.

23. Ø ^* = { ε }。

    星号（Kleene闭包）将某种语言中任意数量的单词聚集在一起以生成结果单词。如果语言为空，则此操作只能收集​​0 个单词，导致仅得到一个空单词。

24. Ø ^* = { ε }.

    The asterisk (Kleene 's closure ) gathers together any number of words from a language to produce a result word. If the language is empty, this operation can only gather 0 words, producing only the empty word.

如果我们假设R是任意正则表达式，我们将获得以下恒等式。它们是对我们是否正确理解定义的考验。

If we assume that R is any regular expression, we obtain the following identities. They are a test of whether we understand the definition correctly.

R ∪ Ø = R。

R ∪ Ø = R .

将空语言添加到任何其他语言不会改变它。

Adding an empty language to any other language will not change it.

R ⚬ ε = R。

R ⚬ ε = R .

在任何单词后添加空格不会改变它。

Appending an empty word to any word will not change it.

然而，如果我们交换前面恒等式中的Ø和ε，可能会导致不再满足等式。

However, if we swapped Ø and ε in the preceding identities, this would cause the equalities to no longer hold.

R ∪ ε可能不等于R。

R ∪ ε may not be equal to R .

例如，如果R = 0，则L ( R ) = {0}，但L ( R ∪ ε ) = { 0 , ε }。

For example, if R = 0 , then L ( R ) = { 0 }, but L ( R ∪ ε ) = { 0 , ε }.

R ⚬ Ø可能不等于R。

R ⚬ Ø may not be equal to R .

例如，如果R = 0，则L ( R ) = {0}，但L ( R ⚬ Ø ) = Ø。

For example, if R = 0 , then L ( R ) = { 0 }, but L ( R ⚬ Ø ) = Ø .

正则表达式是设计编程语言编译器的有用工具。编程语言的基本语法元素称为标记，例如变量名和常量，可以使用正则表达式进行描述。例如，可能包含小数部分和/或符号的数值常量可以被描述为语言元素

Regular expressions are useful tools in the design of programming language compilers. The basic syntactic elements of a programming language, called tokens , such as variable names and constants, can be described using regular expressions. For example, a numeric constant, which may contain a fractional part and/or a sign, can be described as a language element

( + ∪ – ∪ ε ) ( D ⁺ ∪ D ⁺ .D ^* ∪ D ^* .D + ⁾

(+ ∪ – ∪ ε ) ( D ⁺ ∪ D ⁺ . D ^* ∪ D ^* . D ⁺ )

其中D = { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } 是十进制数字的字母表。生成的单词的示例有：72、3.14159、+7。或-.01。

where D = { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } is the alphabet of decimal digits. Examples of generated words are: 72 , 3.14159 , +7. or -.01 .

一旦使用正则表达式通过其标记来描述编程语言的语法，自动化系统就可以生成词法分析器，这是编译器中执行输入程序预处理的部分。

Once the syntax of a programming language is described using regular expressions through its tokens, automated systems can generate a lexical analyzer, a part of the compiler that performs preprocessing of the input program.

与有限自动机的等价

Equivalence with finite automata

正则表达式和有限自动机在表达能力方面是等效的。这个事实令人惊讶，因为有限自动机似乎是完备不同的东西。但是，任何正则表达式都可以转换为可识别表达式所描述的语言的有限自动机，反之亦然。回想一下，正则语言是被某种有限自动机识别的语言。

Regular expressions and finite automata are equivalent in terms of their expressive power. This is surprising, since finite automata seem to be quite different. However, any regular expression can be converted into a finite automaton that recognizes the language described by the expression, and vice versa. Recall that a regular language is one that is recognized by a finite automaton.

定理1.54

Theorem 1.54

当且仅当一种语言由某种正则表达式描述时，它才是正则语言。

A language is regular if and only if it is described by some regular expression.

该定理有两个方向。我们将制定并证明每个方向作为一个单独的引理。

The theorem has two directions. We will formulate and prove each direction as a separate Lemma.

引理 1.55

Lemma 1.55

如果一种语言是用正则表达式描述的，那么它就是正则语言。

If a language is described by a regular expression, it is a regular language.

证明的概念 让我们假设我们有一个描述某种语言A 的正则表达式R。我们将向您展示如何将它们转换为可识别A 的NFA。根据推论 1.40，如果 NFA 识别语言A，则语言A是正则语言。

Concept of Proof Suppose we have a regular expression R describing some language A . We show how to convert it into an NFA that recognizes A . By Corollary 1.40, if the NFA recognizes A , then A is regular.

证明我们将把R转换为 NFA N。我们将考虑正则表达式的正式定义中出现的六种情况。

Proof We convert R to an NFA N . We consider six cases appearing in the formal definition of regular expressions.

1. R = a对于某些a ∈ Σ。那么L ( R ) = { a } 并且以下 NFA 识别L ( R )。

   

   请注意，该机器符合 NFA 定义，但不符合 DFA 定义，因为它包含一些状态，对于每个可能的输入符号都没有传出转换。当然，我们可以在这里提出一个等价的 DFA，但 NFA 对我们来说就足够了，而且更容易描述。

   形式上，N = ({ q ₁ , q ₂ }, Σ , δ , q ₁ , { q ₂ })，其中我们可以通过表述δ ( q ₁ , a ) = { q ₂ } 和δ ( r , b ) = Ø对于r ≠ q ₁或b ≠ a。

2. R = a for some a ∈ Σ . Then L ( R ) = { a }, and the following NFA recognizes L ( R ).

   Note that this machine fits the definition of an NFA but not the definition of a DFA, since it contains some states with no outgoing transitions for each possible input symbol. Of course, we could represent an equivalent DFA here, but an NFA suffices for us and is easier to describe.

   Formally, N = ({ q ₁ , q ₂ }, Σ , δ , q ₁ , { q ₂ }), where δ can be described by stating that δ ( q ₁ , a ) = { q ₂ } and δ ( r , b ) = Ø for r ≠ q ₁ or b ≠ a .

3. R = ε。那么L ( R ) = { ε } 并且以下 NFA 识别L ( R )。

   

   形式上，N = ({ q ₁ }, Σ , δ , q ₁ , { q ₁ })，其中对于任何r和b ， δ ( r , b ) = Ø。

4. R = ε . Then L ( R ) = { ε } and the following NFA recognizes L ( R ).

   Formally, N = ({ q ₁ }, Σ , δ , q ₁ , { q ₁ }), where δ ( r , b ) = Ø for any r and b .

5. R = Ø。那么L ( R ) = Ø并且以下 NFA 识别L ( R )。

   

   形式上，N = ({ q }, Σ , δ , q , Ø )，其中对于任何r和b ， δ ( r , b ) = Ø。

6. R = Ø . Then L ( R ) = Ø and the following NFA recognizes L ( R ).

   Formally, N = ({ q }, Σ , δ , q , Ø ), where δ ( r , b ) = Ø for any r and b .

7. R = R ₁ ∪ R ₂。
8. R = R ₁ ∪ R ₂ .
9. R = R ₁ ⚬ R ₂。
10. R = R ₁ ⚬ R ₂ .
11. R = R ^* ₁
12. R = R ^* ₁

对于最后三种情况，我们将使用证明中给出的构造，即正则语言的类对于正则操作是封闭的。换句话说，我们将使用来自正则语言类封闭性证明的适当构造，从R ₁和R ₂（或在 6 的情况下仅R _{1 ）的 NFA 构造}R的 NFA这个操作。

For the last three cases we will use the constructions given in the proofs that the class of regular languages ​​is closed under regular operations. In other words, we will construct an NFA for R from the NFA for R ₁ and R ₂ (or just R ₁ in case 6) using the appropriate construction from the proof of the closure of the class of regular languages ​​for this operation.

这就完成了定理 1.54 证明的第一部分，确保条件的更简单方向为真当且仅当。在继续第二个方向之前，让我们考虑几个使用上述过程将正则表达式转换为 NFA 的示例。

This completes the first part of the proof of Theorem 1.54, ensuring that the simpler direction of the condition is true if and only if. Before moving to the second direction, let us consider a few examples in which we use the above procedure to transform a regular expression into an NFA.

例 1.56

Example 1.56

我们将通过一系列步骤将正则表达式 ( ab ∪ a ) ^*转换为 NFA。我们将使用自上而下的方法，从最小的子表达式开始，一直到更大的子表达式，直到获得原始表达式的 NFA，如下图所示。请注意，此过程通常不会生成状态最少的 NFA。在此示例中，该过程创建具有八个状态的 NFA，但最小的等效 NFA 仅具有两个状态。你能找到吗？

We will convert a regular expression ( ab ∪ a ) ^* to an NFA in a sequence of steps. We will use a bottom-up approach, starting with the smallest subexpressions and working our way up until we obtain an NFA for the original expression, as shown in the diagram below. Note that this procedure does not generally produce an NFA with the smallest number of states. In this example, the procedure produces an NFA with eight states, but the smallest equivalent NFA has only two states. Can you find it?

图 1.57

从正则表达式构建 NFA ( ab ∪ a ) ^*

Figure 1.57

Building an NFA from a regular expression ( ab ∪ a ) ^*

例 1.58

Example 1.58

图 1.59 显示了正则表达式 ( a ∪ b ) ^* aba到 NFA 的转换。这里不介绍一些小步骤。

Figure 1.59 shows the conversion of a regular expression ( a ∪ b ) ^* aba to NFA. Several minor steps are not shown here.

图 1.59从正则表达式 ( a ∪ b ) ^* aba

构建 NFA

Figure 1.59

Building an NFA from a regular expression ( a ∪ b ) ^* aba

现在让我们从另一个方向来证明定理1.54。

Let us now turn to the proof of Theorem 1.54 in the other direction.

引理 1.60

Lemma 1.60

如果一种语言是正则语言，则它可以表示为正则表达式。它是通过正则表达式来描述的。

If a language is regular, then it can be represented as a regular expression. It is described by a regular expression.

证明的概念我们要证明如果语言A是正则的，那么它可以表示为正则表达式。由于A是正则的，因此它被一些 DFA 接受。我们将描述将确定性有限自动机转换为等效正则表达式的过程。

Proof Concept We are to show that if a language A is regular, then it can be represented as a regular expression. Since A is regular, then it is accepted by some DFA. We describe a procedure for converting deterministic finite automata to equivalent regular expressions.

我们将这个过程分为两部分，使用一种称为广义非确定性有限自动机（GNFA）的新型有限自动机。我们将首先向您展示如何将 DFA 转换为等效的 GNFA，然后如何将 GNFA 转换为正则表达式。

We will split this procedure into two parts, using a new type of finite automaton called a generalized nondeterministic finite automaton ( GNFA). First, we will show how to transform DFAs into equivalent GNFAs, and then how to transform GNFAs into regular expressions.

广义非确定性有限自动机只是非确定性有限自动机，其中转移可以使用任何正则表达式作为标签，而不仅仅是字母表或ε的元素。 GNFA 从输入中读取符号块，不一定像正则 NFA 那样只是单个符号。 GNFA 通过从输入中读取符号块，沿着转换箭头从一种状态移动到另一种状态，该符号块本身形成由该转换的正则表达式描述的单词。 GNFA 是不确定的，因此可以包含处理同一输入单词的许多不同方式。如果通过以这种方式操作，GNFA 可以在到达接受状态时读取整个输入字，则它接受其输入。下图显示了 GNFA 的示例。

Generalized nondeterministic finite automata are simply nondeterministic finite automata in which the transitions can have arbitrary regular expressions as labels, instead of just elements of the alphabet or ε . A GNFA reads blocks of symbols from its input, not necessarily just single symbols like an ordinary NFA. A GNFA transitions from one state to another along a transition arrow by reading a block of symbols from the input, which itself forms a word described by the regular expression for that transition. A GNFA is nondeterministic, and thus can contain many different ways of processing the same input word. A GNFA accepts its input if, acting in this way, it can read the entire input word while still arriving at an accepting state. The figure below shows an example of a GNFA.

图 1.61

广义非确定性有限状态机

Figure 1.61

Generalized nondeterministic finite automaton

为了方便起见，我们将要求 GNFA 始终具有满足以下条件的特殊形式。

For convenience, we will require that a GNFA always have a special form that meets the conditions below.

• 初始状态可以转换到所有其他状态，但没有来自任何其他状态的转换。
• The initial state has transitions to all other states, but no transitions coming from any other state.
• 只能有一个接受状态，并且必须有从所有其他状态到该状态的转换，但不能有到任何其他状态的传出转换。而且，接受状态不能与初始状态相同。
• There can be only one accepting state, and there must be transitions to it from every other state, but no outgoing transitions to any of the other states. Furthermore, the accepting state cannot be identical to the initial state.
• 除了初始状态和接受状态之外，必须存在从每个状态到每个状态的转换，包括到相同状态的转换。
• Except for the start and accepting states, there must be transitions from every state to every state, including those to the same state.

您可以轻松地将任何 DFA 转换为特殊形式的 GNFA。我们只需将具有ε转换的新初始状态添加到旧初始状态，并从所有旧接受状态添加具有ε转换的新接受状态。如果任何转换具有多个标签（或者如果同一方向上的两个状态之间存在多个转换），我们将每个转换替换为一个标签为先前标签之和的单个转换。最后，我们在那些未连接的状态之间添加带有标签Ø的转换。最后一步不会更改机器识别的语言，因为带有标签Ø的转换将永远无法使用。从现在开始，我们将假设所有考虑的 GNFA 都具有特殊形式。

It is easy to transform any DFA into a special-form GNFA. We simply add a new initial state with an ε -transition to the old initial state, and a new accepting state with ε -transitions from all the old accepting states. If any transition has multiple labels (or if there are multiple transitions between two states in the same direction), we replace each with a single transition whose label is the sum of the previous labels. Finally, we add transitions labeled Ø between those states that were not connected. This last step does not change the language recognized by the automaton, since a transition labeled Ø can never be used. From now on, we will assume that all GNFAs under consideration have a special form.

现在我们可以向您展示如何将 GNFA 转换为正则表达式。假设 GNFA 包含k 个状态。然后，由于 GNFA 必须包含初始状态和接受状态，并且它们必须彼此不同，因此我们知道k ≥ 2。如果k > 2，我们将构造一个具有k − 1 个状态的等效 GNFA 。可以在新的 GNFA 上重复此步骤，直到它减少到两个状态。如果k = 2，则 GNFA 包含从初始状态到接受状态的一次转换。此转换的标签是与原始 GNFA 等效的正则表达式。将三态 DFA 转换为等效正则表达式的步骤如下图所示。

Now we can show how to convert a GNFA to a regular expression. Suppose that the GNFA contains k states. Then, since the GNFA must contain initial and accepting states and they must be different, we know that k ≥ 2. If k > 2, then we construct an equivalent GNFA with k − 1 states. This step can be repeated for the new GNFA until it is reduced to two states. If k = 2, then the GNFA contains one transition from the initial state to the accepting state. The label of this transition is a regular expression equivalent to the original GNFA. The steps for converting a three-state DFA to an equivalent regular expression are shown in the figure below.

图 1.62

DFA转换为正则表达式的常见状态

Figure 1.62

Typical states of converting a DFA to a regular expression

关键步骤是当k > 2时构造一个少一个状态的等效 GNFA 。我们通过选择某个状态，将其从机器中删除，然后修复其余状态以便仍然识别相同的语言来实现这一点。任何状态都适合于此，只要它不是初始状态或接受状态。我们确信这样的状态存在，因为k > 2。我们将被删除的状态称为q _rip。

The key step is to construct an equivalent GNFA with one less state when k > 2. We achieve this by choosing some state, removing it from the machine, and repairing the remainder so that the same language is still recognized. Any state will do, as long as it is neither the initial nor the accepting state. We are sure that such a state exists, since k > 2. Let us call the removed state q _rip .

删除q _rip后，我们通过更改每个剩余箭头的标签中的正则表达式来修复自动机。新标签通过添加回丢失的计算来补偿q _rip的缺失。从状态q i到状态q j 的新标签是一个正则表达式，描述可以直接或通过q _rip将机器从状态q i带到q j的所有单词。这种方法如图 1.63 所示。

After removing q _rip , we repair the machine by changing the regular expressions in the labels of each of the remaining arrows. The new labels compensate for the absence of q _rip by adding back the lost computations. The new label leading from state q i to state q j is a regular expression describing all the words that could take the machine from state q i to q j either directly or via q _rip . This approach is illustrated in Figure 1.63.

图 1.63

构建一个少一个状态的等效 GNFA

Figure 1.63

Constructing an equivalent GNFA with one less state

如果在较早的机器上

If in the previous machine

1. q_并导致q_{撕裂带有标签}R ₁的转换，
2. q _and leads to q _rip via the transition labeled R ₁ ,
3. q _{rip通过带有标签}R ₂的转换通向自身，
4. q _rip leads to itself via a transition labeled R ₂ ,
5. q _{rip通过带有标签}R ₃的转换通向q _j，
6. q _rip leads to q _j via a transition labeled R ₃ ,
7. q _i通过带有标签R ₄的转换通向q _j，
8. q _i leads to q _j via a transition labeled R ₄ ,

那么在新机器中，从q _i 到q _{j 的}过渡将被赋予一个标签

then in the new machine the transition from q _i to q _j will be labeled

( R ₁ )( R ₂ ) ^* ( R ₃ ) ∪ ( R ₄ )。

( R ₁ )( R ₂ ) ^* ( R ₃ ) ∪ ( R ₄ ).

我们对从任何状态q _i到任何状态的每个转换进行这样的更改，包括q _i = q _j的情况。新机器可以识别原始语言。

We make this change for each transition from any state q _i to any state, including the case where q _i = q _j . The new machine recognizes the original language.

证明 我们现在将形式化所提出的证明想法。首先，为了便于证明，我们正式定义了上面所示的新型自动机。 GNFA 接近非确定性有限状态机，但转换函数除外，其形式为

Proof We now formalize the presented idea of ​​proof. First, to facilitate the proof, we formally define the new type of automaton shown above. GNFA is similar to a nondeterministic finite automaton, except for the transition function, which has the form

δ : ( Q − { q_接受}) × ( Q − { q_开始}) → R。

δ : ( Q − { q _accept }) × ( Q − { q _start }) → R .

符号R表示字母表Σ上所有正则表达式的集合，q _start和q _Accept分别是初始状态和接受状态。如果δ ( q _i , q _j ) = R ，这意味着从状态q _i到状态q _j的转换以正则表达式R作为其标签。转换函数的域是 ( Q − { q _Accept }) × ( Q − { q _start })，因为根据特殊形式，转换将每个状态连接到每个其他状态，除了没有离开或进入q 的转换之外_接受q_开始。

The symbol R denotes the set of all regular expressions over the alphabet Σ , and q _start and q _accept are the initial and accepting states, respectively. If δ ( q _i , q _j ) = R , this means that the transition from state q _i to state q _j has as its label the regular expression R . The domain of the transition function is ( Q − { q _accept }) × ( Q − { q _start }), since, according to the special form, transitions connect every state to every other state, with the exception that there are no transitions leaving q _accept or entering q _start .

如果w = w ₁ w ₂ …w _k ，则 GNFA 接受属于Σ ^*的词w，其中每个w _i属于Σ ^*，并且存在状态序列q ₀ , q ₁ , …, q _k，使得

GNFA accepts a word w belonging to Σ ^* , if w = w ₁ w ₂ …w _k , where each w _i belongs to Σ ^* , and there is a sequence of states q ₀ , q ₁ , …, q _k , such that

1. q ₀ = q _start是初始状态，
2. q ₀ = q _start is the initial state,
3. q _k = q _accept是接受状态，
4. q _k = q _accept is the accepting state,
5. 对于每个i存在w _i ∈ L ( R _i )，其中R _i = δ ( q _{i −1} , q _i )；换句话说，R _i是描述从q _{i -1}到q _{i 的}转变的表达式。
6. for every i there exists w _i ∈ L ( R _i ), where R _i = δ ( q _{i −1} , q _i ); in other words, R _i is an expression describing the transition from q _{i −1} to q _i .

回到引理 1.60 的证明，令M为语言A的 DFA 。然后，我们可以通过添加新的起始状态和新的接受状态以及必要时的其他转换，将M转换为 GNFA G。我们将使用 CONVERT (G)过程，它接受 GNFA 并返回等效的正则表达式。此过程使用递归，这意味着它会调用自身。我们避免陷入无限循环，因为该过程仅调用自身来处理 GNFA，而 GNFA 的状态少了一个。 GNFA 只有两个状态的情况无需递归处理。

Returning to the proof of Lemma 1.60, let M be a DFA for language A . We can then convert M to a GNFA G by adding a new start state and a new accepting state, and additional transitions as needed. We use the procedure CONVERT (G) , which takes a GNFA and returns an equivalent regular expression. This procedure uses recursion , which means that it calls itself. We avoid getting into an infinite loop because the procedure only calls itself to process a GNFA that has one fewer state. The case where a GNFA has only two states is handled without recursion.

转换（G）：

CONVERT( G ):

1. 令k为机器G的状态数。
2. Let k be the number of states of the automaton G .
3. 如果k = 2，则G必须由初始状态、接受状态以及将它们与包含正则表达式R的标签连接起来的单个转换组成。我们返回表达式R。
4. If k = 2, then G must consist of an initial state, an accepting state, and a single transition connecting them with a label containing the regular expression R . We return the expression R .
5. 当k > 2 时，我们选择与q _start和q _Accept不同的任何状态q _rip ∈ Q，并令G'为 GNFA ( Q' , Σ , δ' , q _start , q _Accept )，其中

   Q' = Q − { q_撕裂},

   对于任意q _i ∈ Q ₀ − { q _Accept } 和任意q _j ∈ Q' − { q _start }，令

   δ' ( q _i , q _j ) = ( R ₁ )( R ₂ ) ^* ( R ₃ ) ∪ ( R ₄ )

   其中，R ₁ = δ ( q _i , q _rip )，R ₂ = δ ( q _rip , q _rip )，R ₃ = δ ( q _rip , q _j ) 且R ₄ = δ ( q _i , q _j )。

6. When k > 2, we choose an arbitrary state q _rip ∈ Q different from q _start and q _accept and let G' be GNFA ( Q' , Σ , δ' , q _start , q _accept ), where

   Q' = Q − { q _rip },

   and for any q _i ∈ Q ₀ − { q _accept } and any q _j ∈ Q' − { q _start }, let

   δ' ( q _i , q _j ) = ( R ₁ )( R ₂ ) ^* ( R ₃ ) ∪ ( R ₄ )

   where R ₁ = δ ( q _i , q _rip ), R ₂ = δ ( q _rip , q _rip ), R ₃ = δ ( q _rip , q _j ) and R ₄ = δ ( q _i , q _j ).

7. 我们计算 CONVERT( G' ) 并返回该值。

   我们现在将证明 CONVERT 返回正确的值。

8. We calculate CONVERT( G' ) and return that value.

   We will now demonstrate that CONVERT returns the correct value.

声明 1.65

Statement 1.65

对于任何 GNFA G，正则表达式CONVERT( G ) 等价于G。

For any GNFA G , the regular expression CONVERT( G ) is equivalent to G .

我们通过对GNFA 状态数k进行归纳来证明这一论点。

We will prove this theorem by induction on k , the number of GNFA states.

基本步骤：我们在k = 2 个状态下证明论文的正确性。如果G只有两个状态，那么它只能包含从初始状态到接受状态的单个转换。这个转换的标签中的正则表达式描述了所有允许G机到达接受状态的单词。因此，这个表达式等价于G。

Basic Step : We prove the theorem for k = 2 states. If G has only two states, then it can contain only a single transition from the initial state to an accepting state. The regular expression in the label of this transition describes all the words that allow G to reach an accepting state. This expression is therefore equivalent to G .

归纳步骤：我们假设该命题对于k − 1 成立，并且我们使用这个假设来证明它对于k状态也成立。首先，我们证明G和G'识别相同的语言。假设G接受中的输入单词。然后，在计算的接受分支中， G一一遍历状态：

Inductive Step : We assume that the thesis is true for k − 1, and we use this assumption to show that it is also true for k states. We first show that G and G' recognize the same language. Suppose G accepts an input word w . Then, in the accepting branch of the computation, G goes through the states in turn:

q_开始，q ₁，q ₂，q ₃，...，q_接受。

q _start , q ₁ , q ₂ , q ₃ , …, q _accept .

如果这些状态都不是q _rip的删除状态，那么显然G'也接受w。这是因为描述G'中转换的每个新正则表达式都包含旧正则表达式作为总和的一部分。

If none of these states is a deleted state of q _rip , then it is obvious that G' also accepts w . This is because each of the new regular expressions describing transitions in G' includes the old regular expressions as part of the union.

如果q _rip出现在上述序列中，则删除所有出现的它会产生一串状态，描述G' 的接受计算。状态q _i和q _j发生在q _rip（状态的子序列）发生之前和之后，对应于连接这些状态的转换上的新正则表达式，它描述了从q _i到q _{j 的}所有单词通过机器G中的q _rip。因此G'接受w。

If q _rip appears in the above sequence, then removing all occurrences of it gives a sequence of states that describes an accepting computation of G' . The states q _i and q _j appearing just before and just after the occurrence of (a subsequence of states of) q _rip correspond to a new regular expression on the transition connecting these states that describes all the words leading from q _i to q _j via q _rip in the automaton G . Thus G' accepts w .

现在考虑相反的方向并假设G'接受中的输入单词。由于G'中任何两个状态q _i和q _j之间的任何转换都直接或通过q _rip描述了G中携带q _i到q _{j 的}一组字，因此G也必须接受w。因此，G和G'是等价的。

Now consider the opposite direction and assume that G' accepts an input word w . Since any transition between any two states q _i and q _j in G' describes the set of words that take q _i to q _j in G either directly or via q _rip , G must also accept w . Hence G and G' are equivalent.

归纳假设指出，当算法针对输入G'递归调用自身时，结果是等价于G'的正则表达式，因为G'包含k − 1 个状态。因此这个正则表达式也等价于G，所以我们证明了算法的正确性。

The inductive conjecture states that when an algorithm calls itself recursively for an input G' , the output is a regular expression that is equivalent to G' , since G' contains k − 1 states. Hence this regular expression is also equivalent to G , and so we have proven the correctness of the algorithm.

这就完成了定理 1.65、引理 1.60 以及定理 1.54 的证明。

This completes the proof of Theorem 1.65, Lemma 1.60, and consequently Theorem 1.54.

实施例 1.66

Example 1.66

在此示例中，我们将使用前面所示的算法将 DFA 转换为正则表达式。我们将从图 1.67(a) 所示的二态 DFA 开始。

In this example, we will use the algorithm shown earlier to transform a DFA into a regular expression. We will start with the two-state DFA shown in Figure 1.67(a).

在图1.67(b)中，我们通过添加一个新的开始状态和一个新的接受状态来创建一个四状态GNFA，分别命名为s和a，而不是q _start和q _accept，这样我们就可以方便地将它们放在数字。为了避免混淆，我们没有用Ø labels 绘制箭头，尽管它们存在。请注意，我们已从输出 DFA 中删除了到状态 2 的环回箭头上的标签a、b，并且标签a ∪ b位于相应的 GNFA 点处。我们这样做是因为 DFA 中的标签代表两种转换，一种用于a，一种用于b，而 GNFA 只能有一种从状态 2 到其自身的转换。

In Figure 1.67(b) we create a four-state GNFA by adding a new start state and a new accepting state, named s and a respectively, instead of q _start and q _accept , so that we can conveniently place them in the figure. To avoid clutter, we have not drawn the arrows labeled Ø , although they are present. Note that we have removed the label a , b on the loopback arrow to state 2 from the output DFA with the label a ∪ b at the corresponding point in the GNFA. We did this because a label in a DFA represents two transitions, one for a and one for b , whereas a GNFA can have only one transition from state 2 to itself.

在图 1.67(c) 中，我们删除了状态 2 并更新了剩余箭头的标签。在这种情况下，唯一会改变的标签是从状态 1 到a 的标签。在 (b) 部分中它是Ø，但在 (c) 部分中它是b ( a ∪ b ) ^*。我们通过执行 CONVERT 过程的步骤 3 获得此结果。状态q _i是状态 1，状态q _j是a，q _rip是状态 2，因此R ₁ = b、R ₂ = a ∪ b、R ₃ = ε和R ₄ = Ø。因此，从状态 1 到a 的箭头的新标签是 (b)( a ∪ b ) ^* ( ε ) ∪ Ø。该正则表达式可以简化为b ( a ∪ b ) ^*。

In Figure 1.67(c), we have removed state 2 and updated the labels of the remaining arrows. In this case, the only label that changes is the one from state 1 to a . In part (b), this was Ø , but in part (c) it is b ( a ∪ b ) ^* . We obtained this result by executing step 3 of the CONVERT procedure. State q _i is state 1, state q _j is a , and q _rip is state 2, so R ₁ = b , R ₂ = a ∪ b , R ₃ = ε , and R ₄ = Ø . Thus, the new label of the arrow from state 1 to a is (b)( a ∪ b ) ^* ( ε ) ∪ Ø . This regular expression can be simplified to b ( a ∪ b ) ^* .

在图 1.67(d) 中，我们从 (c) 部分中删除了状态 1 并执行了类似的过程。由于只剩下初始状态和接受状态，因此连接它们的箭头标签是与原始DFA等效的正则表达式。

In Figure 1.67(d), we have removed state 1 from part (c) and performed an analogous procedure. Since only the initial state and the accepting state remain, the label of the arrow connecting them is a regular expression equivalent to the original DFA.

图 1.67

将二态 DFA 转换为等效的正则表达式

Figure 1.67

Converting a two-state DFA to an equivalent regular expression

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例 1.68

Example 1.68

在此示例中，我们从三态 DFA 开始。后续转换步骤如下图所示。

In this example, we start with a three-state DFA. The conversion steps are shown in the following illustration.

图 1.69

将三态 DFA 转换为等效的正则表达式

Figure 1.69

Converting a three-state DFA to an equivalent regular expression

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1.4

非正则语言

1.4

Irregular languages

要了解有限状态机的强大功能，您还需要了解它们的局限性。在本节中我们将展示如何证明某些语言无法被任何有限自动机识别。

To understand the power of finite automata, one must also understand their limitations. In this section, we will show how to prove that certain languages ​​cannot be recognized by any finite automaton.

让我们假设语言B = { 0 ⁿ 1 ⁿ : n ≥ 0}。如果我们尝试找到一个能够识别语言B的 DFA ，我们会发现机器在读取输入时似乎必须记住到目前为止它已经看到了多少个零。由于这些零的数量没有限制，因此机器必须跟踪无限数量的可能性。然而，这对于任何有限数量的状态都是不可行的。

Assume the language B = { 0 ⁿ 1 ⁿ : n ≥ 0}. If we try to find a DFA that recognizes the language B , we find that the machine, it seems, needs to remember how many zeros it has seen so far while reading the input. Since the number of these zeros is not bounded, the machine will have to keep track of an infinite number of possibilities. However, this is not feasible with any finite number of states.

然后我们提出一种方法来证明像B这样的语言不是正则语言。由于零的数量是无限的，所提出的论点是否已经证明了非正则性？他不领导。仅仅因为一种语言看起来需要无限的内存并不一定意味着它实际上需要无限的内存。这对于语言B来说实际上是正确的；然而，还有其他语言看似也需要无限数量的可能性，尽管实际上它们是正则的。例如，考虑字母表上的两种语言Σ = { 0 , 1 }：

Next, we will present a method for proving that languages ​​like B are not regular. Doesn't the argument already prove irregularity, since the number of zeros is unlimited? It doesn't. Just because a language seems to require unlimited memory doesn't mean that it does. This is indeed true for B ; but there are other languages ​​that also seem to require an unlimited number of possibilities, but are in fact regular. For example, consider two languages ​​over the alphabet Σ = { 0 , 1 }:

C = { w : w包含相同数量的 0 和 1}，

C = { w : w contains an equal number of zeros and ones},

D = { w : w包含相同次数的 01和10作为子词}。

D = { w : w contains an equal number of occurrences of 01 and 10 as substrings}.

乍一看，人们可能会认为识别机器必须计算每个实例中的出现次数，因此这些语言中没有一种看起来是规则的。正如预期的那样，C不是正则的，但令人惊讶的是，D是正则的！⁶因此，直觉有时会让我们误入歧途，这就是为什么我们需要数学证据来确定。在本节中我们将向您展示如何证明某些特定语言是非正则的。

At first glance, one might think that the recognition machine must count the occurrences in each case, and thus none of these languages ​​seem regular. As expected, C is not regular, but surprisingly, D is regular! ⁶ So intuition can sometimes lead us astray, and that is why we need mathematical proofs to be sure. In this section, we show how to prove that certain specific languages ​​are not regular.

正则语言的泵送引理

The pumping lemma for regular languages

我们证明非正则性的技术源自正则语言定理，传统上称为泵引理。该定理指出所有正则语言都具有一些特殊属性。如果我们能够证明某种语言不具有这种性质，我们就可以确定它不是正则的。该属性表明，语言中的任何单词都可以“膨胀”，只要它至少与某个特定值（称为泵长度）一样长。我们的意思是，这样的单词包含可以重复任意多次的部分（子词），并且生成的单词仍然属于该语言。

Our technique for proving irregularity comes from a theorem about regular languages, traditionally called the pumping lemma . This theorem states that all regular languages ​​have a certain special property. If we can show that a language does not have this property, we are certain that it is not regular. This property states that any word of the language can be "pumped up" if it is at least as long as some special value, called the pumping length . By this we mean that such a word contains a part (a subword) that can be replicated any number of times, and the resulting word will still belong to the language.

定理1.70

Theorem 1.70

泵 引理 如果A是正则语言，则存在一个数字p（泵长度），使得如果s是A中长度至少为p的任何单词，则s可以分为三部分， s = xyz，满足符合以下条件： :

Pumping  Lemma  If A is a regular language, then there is a number p (the pumping length) such that if s is any word in A of length at least p , then s can be split into three parts, s = xyz , satisfying the following conditions:

1. 对于所有i ≥ 0，xyⁱz ∈ A，
2. for all i ≥ 0, xy ⁱ z ∈ A ,
3. | y | > 0,
4. | y | > 0,
5. | XY | ≤ p。
6. | xy | ≤ p .

让我们回想一下条目 | s |是单词s的长度，y ⁱ是单词y的i 个副本的串联，并且y ⁰等于ε。

Recall that the notation | s | denotes the length of word s , y ⁱ is the concatenation of i copies of word y , and y ⁰ is equal to ε .

当s被划分为xyz部分时，x或z可以是空字( ε )，但条件 2 表明y ≠ ε。可以看出，如果没有条件 2，该陈述就很简单为真。条件 3 规定x和y部分的长度最多为p。这是一个附加的技术条件，有时在证明某些语言非正则时很有用。对于条件3的应用，参见例1.74。

When s is partitioned into parts xyz , either x or z can be empty words ( ε ) , but condition 2 states that y ≠ ε . It can be seen that without condition 2 the theorem would be trivially true. Condition 3 states that the parts x and y together have length at most p . This is an additional technical condition that is sometimes useful in proving that certain languages ​​are not regular. An application of condition 3 is given by Example 1.74.

证明的概念设M =( Q , Σ , δ , q ₁ , F ) 是一个识别A 的DFA 。我们将泵浦长度p指定为等于状态数M的值。我们将证明语言A中长度至少为p的任何单词都可以分为满足我们的三个条件的三个部分xyz。但是如果A中没有一个单词的长度至少为p呢？那么我们的任务就更容易了，因为定理变得微不足道：显然，如果不存在这样的单词，则长度至少为p的所有单词都满足所有三个条件。

Concept of the Proof Let M = ( Q , Σ , δ , q ₁ , F ) be a DFA recognizing A . We assign the pumping length p to a value equal to the number of states in M ​​. We show that any word of length at least p in A can be partitioned into three parts xyz satisfying our three conditions. But what if no word in A has length at least p ? Then our task is even easier, since the theorem becomes trivially true: of course, all three conditions are satisfied for all words of length at least p , if no such words exist.

如果属于A的s的长度至少为p ，请考虑机器M在处理输入s时经过的状态序列。它从初始状态q ₁开始，然后进入状态q ₃，然后进入q ₂₀，然后进入q ₉等等，直到到达状态q _{13中的单词}s的末尾。由于s属于A，我们知道M接受s，因此q ₁₃是接受状态。

If s of A is of length at least p , consider the sequence of states that machine M goes through while processing input s . It starts with the initial state q ₁ , then goes to, for example, state q ₃ , then to q ₂₀ , then to q ₉ , and so on, until it reaches the end of word s in state q ₁₃ . Since s is of A , we know that M accepts s , so q ₁₃ is an accepting state.

如果我们假设n是s的长度，则状态序列q ₁ , q ₃ , q ₂₀ , q ₉ , …, q _{13 的}长度为n + 1。由于n不小于p，所以我们知道n + 1 大于p ，即状态数M。因此，字符串必须包含重复状态。这个结果是鸽巢规则的一个例子，这是一个有点奇特的名称，用于描述一个相当明显的事实：如果我们尝试将p 个项目排列到少于p 个盒子中，则某个盒子必须包含多个项目。

If we take n to be the length of s , then the sequence of states q ₁ , q ₃ , q ₂₀ , q ₉ , …, q ₁₃ has length n + 1. Since n is not less than p , we know that n + 1 is greater than p , the number of states of M . Thus, the sequence must contain a repeated state. This result is an example of the pigeonhole principle , a slightly fancy name for the rather obvious fact that if we try to arrange p items into fewer than p pigeonholes, some pigeonhole must contain more than one item.

下图显示了单词s以及M在处理s时经历的状态序列。在此示例中，状态q ₉是重复的状态。

The following figure shows a word s and the sequence of states that M goes through while processing s . In this example, state q ₉ is the one that repeats.

图 1.71 M读取单词s时

的重复状态示例

Figure 1.71

Example showing the repeating state when reading word s by M

我们现在可以将s分为三部分x、y和z。x部分是s中出现在q ₉之前的部分，y部分是两次出现q ₉之间的部分，z部分是s中第二次出现q ₉之后的剩余部分。因此，x将M从状态q ₁带到q ₉，y将 M 从q ₉带到q ₉，z将M从q ₉带到接受状态q ₁₃，如下图所示。

We can now divide s into three parts x , y , and z . The x part is the part of s that occurred before q ₉ , the y part is the part between the two occurrences of q ₉ , and the z part is the remaining part of s that occurs after the second occurrence of q ₉ . Thus x takes M from state q ₁ to q ₉ , y takes M from q ₉ back to q ₉ , and z takes M from q ₉ to the accepting state q ₁₃ , as in the figure below.

图 1.72

显示单词x、y和z如何影响M的示例

Figure 1.72

An example showing how the words x , y , and z affect M

让我们看看为什么s 的这种除法满足给定的条件。假设我们运行机器M来输入xyyz。我们知道x将M从状态q ₁带到q ₉，然后第一个y将M从q ₉带回到q _{9 ，第二个}y也是如此，然后z将机器带到状态q ₁₃。由于q ₁₃是接受状态，因此M接受输入xyyz。类似地，对于任何i > 0，它也会接受xy^和z 。在i = 0的情况下，xy^和z = xz，出于类似的原因可以接受。这确保满足条件 1。

Let's see why such a partition s satisfies the given conditions. Suppose we run the machine M for input xyyz . We know that x takes M from state q ₁ to q ₉ , and then the first y takes M from q ₉ back to q ₉ , which does the second y , and then z moves the machine to state q ₁₃ . Since q ₁₃ is an accepting state, M accepts input xyyz . Similarly, it will also accept xy ^and z for any i > 0. For i = 0, xy ^and z = xz , which is accepted for analogous reasons. This satisfies condition 1.

通过检查条件 2，我们可以看到 | y | > 0，因为它是s的一部分，发生在状态q ₉的两次不同出现之间。

Checking condition 2, we can see that | y | > 0, as it is the part of s that occurred between two different occurrences of state q ₉ .

为了获得条件3，我们需要确保q ₉是单词中第一个重复的条件。根据鸽笼规则，序列中的前p +1 个状态必须包含重复。因此| XY | ≤ p。

To obtain condition 3, we need to make sure that q ₉ is the first repeating state in the word. By the pigeonhole principle, the first p + 1 states in the sequence must contain a repeat. Hence, | xy | ≤ p .

证明 设M = ( Q , Σ , δ , q ₁ , F ) 为识别语言A 的DFA，设p为M 的状态数。

Proof Let M = ( Q , Σ , δ , q ₁ , F ) be a DFA recognizing language A and let p be the number of states of M .

设s = s ₁ s ₂ …s _n是属于A的长度为n的单词，其中n ≥ p。令r ₁ , …, r _{n +1为}M在处理s期间进入的状态序列，因此r _{i +1} = δ ( r _i , s _i )，其中 1 ≤ i ≤ n。该序列的长度为n + 1，即至少p + 1。根据鸽笼规则，在序列r ₁ , ..., r _{n +1的前}p + 1 个元素中，至少有两个状态必须相同。我们将第一个称为r _j，将第二个称为r _l。由于r _l出现在以r ₁开头的单词的前p + 1 个位置之一，因此我们得到l ≤ p + 1。现在令x = s ₁ …s _{j −1}、y = s _j …s _{l −1}且z = s _l …s _n。

Let s = s ₁ s ₂ …s _n be a word of length n in A , where n ≥ p . Let r ₁ , …, r _{n +1} be a sequence of states that M enters while processing s , such that r _{i +1} = δ ( r _i , s _i ) for 1 ≤ i ≤ n . This sequence has length n + 1, that is, at least p + 1. By the pigeonhole principle, among the first p + 1 elements of the sequence r ₁ , …, r _{n +1} at least two states must be identical. Let us call the first of them r _j , and the second r _l . Since r _l occurs in one of the first p +1 positions of the word starting with r ₁ , we get l ≤ p +1. Now let x = s ₁ …s _{j −1} , y = s _j …s _{l −1} and z = s _l …s _n .

由于x将M从状态r ₁转移到r _j，y将M从状态r _j转移到r _j，z从r _j转移到r _{n +1}（这是接受状态），因此当i ≥ 0 时， M必须接受xy^和z。我们知道j ≠ l，所以 | y | > 0；我们还知道l ≤ p +1，所以 | XY | ≤ p。这样，我们就满足了泵送引理的所有条件。

Since x takes M from state r ₁ to r _j , y takes M from r _j to r _j , and z from r _j to r _{n +1} , which is an accepting state, M must accept xy ^and z for i ≥ 0. We know that j ≠ l , so | y | > 0; we also know that l ≤ p +1, so | xy | ≤ p . We have thus satisfied all the conditions of the Pumping Lemma.

为了使用泵引理来证明语言B不是正则的，我们首先假设B是正则的，从而导致矛盾。然后我们使用泵引理，它保证存在泵长度p，使得属于B且长度等于p或更大的每个单词都可以被膨胀。然后我们找到一个长度为p或更大且不能膨胀的单词s 。最后，我们表明，在给定将s划分为x、y和z部分的所有方式的情况下，单词s不能被泵送（如果方便的话，考虑泵送引理的条件 3），并且对于每个这样的分区，我们找到 的值i使得xy^和z ∉ B。最后一步通常涉及将不同的划分方式分组为不同的情况并单独分析它们。s 的存在与泵引理相矛盾，如果语言B是正则的，泵引理就必须成立。由此可见B不可能是正则语言。

To use the Pumping Lemma to show that a language B is not regular, we first assume that B is regular to achieve a contradiction. We then use the Pumping Lemma, which guarantees the existence of a pumping length p such that every word of length p or greater in B can be pumped. We then find a word s of length p or greater that cannot be pumped. Finally, we show that the word s cannot be pumped, given all the ways of partitioning s into parts x , y , and z (taking into account condition 3 of the Pumping Lemma, if convenient), and for each such partition we find a value for i such that x y ^and z ∉ B . This final step often requires grouping the different ways of partitioning s into different cases and analyzing them individually. The existence of s contradicts the Pumping Lemma, which would have to hold if B were regular. It follows that B cannot be a regular language.

寻找s有时需要一些创造性思维。在找到一个有效的候选者之前，您可能需要尝试许多候选者。值得尝试的B元素似乎代表了B非正则性的“本质”。我们将在以下一些示例中进一步讨论搜索任务。

Finding s sometimes requires a bit of creative thinking. You may need to try many candidates for s before you find one that works. It is worth trying out elements of B that seem to represent the “essence” of the irregularity of B . We will discuss the task of finding s further in some of the following examples.

实施例 1.73

Example 1.73

设B为语言 { 0 ⁿ 1 ⁿ : n ≥ 0}。我们将使用泵引理来证明B不是正则语言。通过制造矛盾来证明。

Let B be a language { 0 ⁿ 1 ⁿ : n ≥ 0}. We use the pumping lemma to prove that B is not a regular language. The proof follows by contradiction.

假设B是正则的。令p为由泵浦引理得出的泵浦长度。作为s，我们选择单词 0 ^p 1 ^p。由于s属于B并且长度大于p，泵送引理保证s可以分为三部分，s = xyz，这样对于任何i ≥ 0，单词xy和^z也属于B。我们将考虑三种情况来证明这样的结果是不可能的。

Assume that B is regular. Let p be the pumping length that follows from the Pumping Lemma. We choose s to be the word 0 ^p 1 ^p . Since s belongs to B and has length greater than p , the Pumping Lemma guarantees that s can be partitioned into three parts, s = xyz , such that for any i ≥ 0 the words xy ^and z also belong to B . We consider three cases to show that such a result is impossible.

1. 由全零组成的子字y 。在这种情况下，单词xyyz的 0 多于 1，因此不是B 的元素，这违反了泵引理的条件 1。这个案例就导致了一个矛盾。
2. A subword y consisting of all zeros. In this case, the word xyyz has more zeros than ones and thus is not an element of B , which violates condition 1 of the Pumping Lemma. This case leads to a contradiction.
3. 子词y仅由 1 组成 - 类比推理使我们陷入矛盾。
4. The subword y is composed of only ones – analogous reasoning leads us to a contradiction.
5. 子字y由 0 和 1 组成。在这种情况下，单词xyyz可能包含相同数量的 0 和 1，但它们不会被排序，即一些 1 将出现在一些 0 之前。因此，它不是语言B的元素，这意味着矛盾。
6. The subword y consists of both zeros and ones. In this case, the word xyyz can contain the same number of zeros and ones, but they will not be in order, i.e. some ones will come before some zeros. Thus, it is not an element of the language B , which means a contradiction.

由于假设 B 是正则语言会导致矛盾，因此我们得出结论 B 不是正则语言。请注意，我们可以通过使用泵引理的条件 3 来消除情况 2 和 3 来简化该论证。

Since the assumption that B is regular leads to a contradiction, we conclude that B is not a regular language. Note that we can simplify this argument by using condition 3 of the pumping lemma to eliminate cases 2 and 3.

在这个例子中，找到单词s很容易，因为B中任何长度为p或更大的单词都会不一致。在接下来的两个示例中，某些s字选择将不起作用，因此需要付出额外的努力。

In this example, finding s was easy, since any word in B of length p or greater will show the contradiction. In the next two examples, some choices of s will not work, so additional effort is required.

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实施例 1.74

Example 1.74

令C = { w : w包含相同数量的 0 和 1}。我们将使用泵引理来证明C不是正则语言。证明将通过还原为矛盾来进行。

Let C = { w : w contains an equal number of zeros and ones}. We use the pumping lemma to show that C is not a regular language. The proof follows by reduction to contradiction.

相反，假设C是正则的。令p为由泵浦引理得出的泵浦长度。如示例 1.73 所示，令s为单词0 ^p 1 ^p。由于s是C的成员并且长度大于p，泵送引理保证s可以分为三部分，s = xyz，这样对于任何i  ≥  0，单词xy^和z也属于C。我们想要表明这样的结果是不可能的。但是等等，这是可能的！如果我们假设x和z是空词，y是词0 ^p 1 ^p，则xy^和z始终具有相同数量的 0 和 1，因此属于C。所以看起来s可以膨胀。

Suppose, on the contrary, that C is regular. Let p be the pumping length that follows from the Pumping Lemma. As in Example 1.73, let s be the word 0 ^p 1 ^p . Since s is a member of C and has length greater than p , the Pumping Lemma guarantees that s can be partitioned into three parts, s = xyz , such that for any i  ≥  0 , the word xy ^and z also belong to C . We intend to show that such a result is impossible. But wait, it is possible! If we assume that x and z are empty words and y is the word 0 ^p 1 ^p , then xy ^and z always have an equal number of zeros and ones, and thus belong to C . So it looks like s can be pumped.

这就是泵送引理的条件 3 发挥作用的地方。它强制要求在泵送单词s时，必须将其拆分，以便 | XY | ≤ p。对s如何划分的这种限制很容易表明我们选择的单词s = 0 ^p 1 ^p不能被夸大。如果 | XY | ≤ p，这意味着y必须由全零组成，因此xyyz ∉ C。因此，s不能被泵浦，这给我们带来了所需的矛盾。

This is where Condition 3 of the Pumping Lemma comes in handy. It dictates that if a word s is pumped , it must be partitioned so that | xy | ≤ p . This restriction on the way s is partitioned makes it easy to show that our chosen word s = 0 ^p 1 ^p cannot be pumped. If | xy | ≤ p , this means that y must consist of all zeros, so xyyz ∉ C . Hence s cannot be pumped, which gives us the desired contradiction.

在本例中选择s字需要比例 1.73 中更多的努力。例如，如果我们选择s = ( 01 ) ^p，我们可能会遇到麻烦，因为我们需要找到一个不能膨胀的单词，并且即使给定条件 3，这个特定的单词也可以膨胀。你能看到它是如何膨胀的吗？一种方法是设置x = ε、y = 01和z = ( 01 ) ^{p −1}。然后对于每个单词xy^和z ∈ C。如果您第一次未能找到一个无法夸大的单词，请不要绝望，而是尝试另一个单词！

Choosing the word s in this example required more effort than in Example 1.73. For example, if we had chosen s = ( 01 ) ^p , we might have run into trouble, since we need to find a word that cannot be inflated, and this particular word can be inflated even under Condition 3. Can you see how it can be inflated? One way to do this is to set x = ε , y = 01 , and z = ( 01 ) ^{p −1} . Then for each i word xy ^and z ∈ C . If you don't find a word that cannot be inflated on the first try, don't despair, but try another one!

证明C非正则的另一种方法是根据我们对B非正则的知识得出的。如果C是正则的，则C ∩ 0 ^* 1 ^*也将是正则的。这是因为语言0 ^* 1 ^*是正则语言，正则语言的类由于交集而封闭，正如我们在脚注 3（第 46 页）中所示。然而，C ∩ 0 ^* 1 ^*等于B，并且我们从例 1.73 中已经知道B不是正则的。

An alternative way of proving that C is not regular follows from our knowledge that B is not regular. If C were regular, C ∩ 0 ^* 1 ^* would also be regular. This follows from the fact that the language 0 ^* 1 ^* is regular and the class of regular languages ​​is closed under intersection, as we showed in footnote 3 (page 46). However, C ∩ 0 ^* 1 ^* is equal to B , and we already know from Example 1.73 that B is not regular.

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实施例 1.75

Example 1.75

令F = { ww : w ∈ { 0 , 1 } ^* }。我们将使用泵引理证明F不是正则语言。

Let F = { ww : w ∈ { 0 , 1 } ^* }. We will show that F is not a regular language using the pumping lemma.

为了证明的目的，我们隐含地假设F是正则的。令p为由泵浦引理得出的泵浦长度。令s为单词0 ^p 10 ^p 1。由于s属于F并且s的长度大于p，因此泵送引理保证s可以分为三部分，s = xyz，满足引理的三个条件。我们将证明这样的结果是不可能的。

For the sake of proof, we assume that F is regular. Let p be the pumping length that follows from the Pumping Lemma. Let s be the word 0 ^p 10 ^p 1 . Since s is a member of F and s has length greater than p , the Pumping Lemma guarantees that s can be partitioned into three parts, s = xyz , satisfying the three conditions of the lemma. We will show that such a result is impossible.

条件 3 再次被证明是至关重要的，因为没有它我们可以泵 s，假设x和z是空字。由于条件 3，证明与前一个类似，因为y必须仅由零组成，因此xyyz ∉ F。

Condition 3 is once again crucial, since without it we could pump  s by assuming that x and z are empty words. Thanks to Condition 3, the proof is analogous to the previous one, since y must consist of all zeros, therefore xyyz ∉ F .

请注意，我们选择s = 0 ^p 10 ^p 1作为代表非正则性F的“本质”的单词，而不是例如单词0 ^p 0 ^p。即使0 ^p 0 ^p属于F，也不可能证明矛盾，因为它可以被泵浦。

Note that we have chosen s = 0 ^p 10 ^p 1 as the word representing the “essence” of the irregularity F , in contrast to, for example, the word 0 ^p 0 ^p . Even though 0 ^p 0 ^p belongs to F , we will not be able to demonstrate a contradiction with it, since it can be pumped up.

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实施例 1.76

Example 1.76

我们将展示一种非正则的一元语言。令D = { 1 ^{n 2} : n ≥ 0}。换句话说，D包含所有由长度为自然数平方的单词组成的单词。我们将使用泵引理来证明D不是正则语言。通过制造矛盾来证明。

We will show an irregular unary language. Let D = { 1 ^{n 2} : n ≥ 0}. In other words, D contains all words built of ones whose lengths are squares of natural numbers. We will use the pumping lemma to show that D is not a regular language. The proof follows by contradiction.

假设D是正则语言。令p为由泵浦引理得出的泵浦长度。让s成为单词1 ^{p 2}。由于s是D的元素，并且s的长度至少为p，因此泵送引理保证s可以分为三部分，s = xyz，满足引理的三个条件。与前面的示例一样，我们将证明这样的结果是不可能的。这需要对自然数的平方序列进行思考：

Assume D is a regular language. Let p be the pumping length that follows from the Pumping Lemma. Let s be a word 1 ^{p 2} . Since s is an element of D and the length of s is at least p , the Pumping Lemma guarantees that s can be partitioned into three parts, s = xyz , satisfying the three conditions of the lemma. As in the previous examples, we will show that such a result is impossible. This requires a moment's thought about the sequence of squares of natural numbers:

0, 1, 4, 9, 16, 25, 36, 49, …

0, 1, 4, 9, 16, 25, 36, 49, …

让我们注意该序列的后续元素之间不断增加的差异。该序列的大元素不能彼此靠近。

Notice the increasing differences between successive elements of this sequence. Large elements of this sequence cannot be close to each other.

现在考虑两个单词xyz和xy ² z。这些单词因y 的一次重复而彼此不同，因此它们的长度也因y的长度而不同。从泵送引理的条件 3 | XY | ≤ p，因此 | y | ≤ p。所以我们有 | XYZ | = p ²和 | XY ² Z | ≤ p ² + p。但p ² + p < p ² +2 p +1 = ( p +1) ²。此外，条件 2 意味着y不能是空词，因此 | XY ² Z | >第²页。因此，字长xy ² z位于两个连续的方格p ²和( p +1) ²之间。因此，长度本身不可能是整数的平方。因此，我们得出了xy ²与∉ D 的矛盾，并得出语言D不是正则的结论。

Now consider two words xyz and xy ² z . These words differ from each other by a single repetition y , and consequently their lengths differ by length y . By condition 3 of the Pumping Lemma, | xy | ≤ p , and therefore | y | ≤ p . We thus have | xyz | = p ² and | xy ² z | ≤ p ² + p . But p ² + p < p ² +2 p +1 = ( p +1) ² . Moreover, by condition 2 it follows that y cannot be the empty word, and therefore | xy ² z | > p ² . Thus the length of the word xy ² z lies between two consecutive squares p ² and ( p + 1) ² . Hence this length cannot itself be an integer square. We have thus reached the contradiction xy ² z ∉ D and we conclude that the language D is not regular.

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例 1.77

Example 1.77

有时，在应用抽吸引理时，“抽吸”很有用。我们将使用泵引理来证明语言E = { 0 ⁱ 1 ^j : i > j } 是非正则的。通过制造矛盾来证明。

Sometimes it is useful to "pump down" when applying the Pumping Lemma. We will use the Pumping Lemma to show that the language E = { 0 ⁱ 1 ^j : i > j } is not regular. The proof follows by contradiction.

假设E是正则语言。令p为由泵浦引理得出的泵浦长度E。令s = 0 ^{p +1} 1 ^p。然后s可以分为单词xyz，满足引理的条件。根据条件 3，y由全零组成。让我们检查一下单词xyyz，看看它是否可以属于E。附加y的额外副本会增加零的数量。然而，E包含属于0 ^* 1 ^*的所有单词，其中包含的 0 多于 1，因此增加 0 的数量仍然会产生属于E 的单词。这里并不矛盾。我们必须尝试一些不同的东西。

Suppose E is a regular language. Let p be the pumping length of E , which follows from the Pumping Lemma. Let s = 0 ^{p +} 1 1 ^p . Then s can be partitioned into words xyz , which satisfy the conditions of the lemma. By Condition 3, y consists of all zeros. Let us examine the word xyyz to see if it can be a member of E . Appending an additional copy of y increases the number of zeros. However, E contains all the words in 0 ^* 1 ^* , which contain more zeros than ones, so increasing the number of zeros still gives a word in E . There is no contradiction here. We have to try something else.

泵引理指出，即使当i = 0 时， xy ⁱ z ∈ E也如此，所以让我们尝试考虑xy ⁰ z = xz。删除单词y会减少s中零的数量。然而，回想一下，s只包含比 1 多 1 个的 0。因此，xz 的0不能多于 1，因此不能属于E。这就是我们如何得到矛盾的原因。

The pumping lemma states that xy ⁱ z ∈ E even when i = 0, so let's consider xy ⁰ z = xz . Removing the term y reduces the number of zeros in s . Recall, however, that s contains only one more zero than ones. Therefore, xz cannot have more zeros than ones, and therefore cannot belong to E . This gives us a contradiction.

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练习

Exercises

^A 1.1下面显示的是两个DFA M ₁和M ₂ 的状态图。回答以下关于这些机器的问题。

^A 1.1 Shown below are state diagrams of two DFAs M ₁ and M ₂ . Answer the following questions about these machines.

a. 初始状态是什么？

a. What is the initial state?

b. 接受状态的集合是什么？

b. What is the set of accepting states?

c.机器在处理aabb输入 时经历什么状态序列？

c. What sequence of states does the machine go through when processing input aabb ?

d. 机器接受单词aabb吗？

d. Does the machine accept the word aabb ?

e. 机器接受单词ε吗？

e. Does the machine accept the word ε ?

^A 1.2给出练习 1.1 中所示 机器M ₁和M _{2的正式描述。}

^A 1.2 Give a formal description of the machines M ₁ and M ₂ presented in exercise 1.1.

1.3 DFA M 的形式化描述为({ q ₁ , q ₂ , q ₃ , q ₄ , q ₅ }, { u , d }, δ , q ₃ , { q ₃ })，其中δ由表定义以下。提供该机器的状态图。

1.3 The formal description of the DFA M is ({ q ₁ , q ₂ , q ₃ , q ₄ , q ₅ }, { u , d }, δ , q ₃ , { q ₃ }), where δ is defined by the table below. Give a state diagram for this machine.

	在 at	d d
q 1 q1​	q 1 q1​	q 2 q2​
q 2 q2​	q 1 q1​	q 3 q3​
q 3 q3​	q 2 q2​	q 4 q4​
q 4 q4​	q 3 q3​	q 5 q5​
q 5 q5​	q 4 q4​	q 5 q5​

1.4 以下每种语言都是两种更简单语言的交集（共同）。在每一点，为更简单的语言构建一个 DFA，然后使用脚注 3（第 46 页）中描述的结构将它们组合起来，以表示给定语言的 DFA 状态图。在所有点Σ = { a , b }。

1.4 Each of the following languages ​​is the intersection of two simpler languages. At each point, construct a DFA for the simpler languages, and then combine them using the construction described in footnote 3 (page 46) to represent a DFA state diagram for the given language. At all points, Σ = { a , b }.

a. { w : w包含至少两个a和至少两个b }

a. { w : w contains at least two a's and at least two b's }

^A b. { w : w恰好包含两个a和至少两个b }

^A b. { w : w contains exactly two a's and at least two b's }

c. { w : w包含一个偶数a和一个或两个b }

c. { w : w contains an even number of a's and one or two b's }

^A d. { w : w包含偶数个a，每个a后面至少有一个b }

^A d. { w : w contains an even number of a , where each a is followed by at least one b }

e. { w : w以a开头，最多包含一个b }

e. { w : w starts with a and contains at most one b }

f. { w : w包含奇数a并以b结尾}

f. { w : w contains an odd number of a's and ends with b }

g. { w : w具有偶数长度并包含奇数a }

g. { w : w has even length and contains an odd number of a }

1.5 以下每种语言都是更简单语言的补充。在每一点，为更简单的语言构建一个 DFA，然后用它来表示给定语言的 DFA 的状态图。在所有点Σ = { a , b }。

1.5 Each of the following languages ​​is the complement of a simpler language. At each point, construct a DFA for the simpler language, and then use it to represent a DFA state diagram for the given language. At all points, Σ = { a , b }.

^A a. { w : w不包含子词ab }

^A a. { w : w does not contain the subword ab }

^A b. { w : w不包含子词baba }

^A b. { w : w does not contain the subword baba }

c. { w : w不包含任何子串ab或ba }

c. { w : w does not contain any of the substrings ab or ba }

d. { w : w是不属于a ^* b ^*的任何单词}

d. { w : w is any word not belonging to a ^* b ^* }

e. { w : w是不属于 ( ab ⁺ ) ^*的任何单词}

e. { w : w is any word not belonging to ( ab ⁺ ) ^* }

f. { w : w是不属于a ^* ∪ b ^*的任何单词}

f. { w : w is any word not belonging to a ^* ∪ b ^* }

g. { w : w是任何不包含两个a 的单词}

g. { w : w is any word that does not contain exactly two a's }

h. { w : w是除a和 b之外的任何单词}

h. { w : w is any word except a and  b }

1.6 提供识别以下语言的 DFA 的状态图。在所有点上，字母表都是 { 0 , 1 }。

1.6 Give state diagrams for DFAs that recognize the following languages. At all points the alphabet is { 0 , 1 }.

a. { w : w以1开头，以0结尾}

a. { w : w starts with 1 and ends with 0 }

b. { w : w至少包含三个1 }

b. { w : w contains at least three 1 }

c. { w : w包含子字0101（即w = x 0101 y对于某些x和y）}

c. { w : w contains the substring 0101 (i.e. w = x 0101 y for some x and y )}

d. { w : w的长度至少为 3 并且其第三个符号为0 }

d. { w : w has length at least 3 and its third symbol is 0 }

e. { w : w从0开始并且具有奇数长度或从1开始并且具有偶数长度}

e. { w : w starts from 0 and has odd length or starts from 1 and has even length}

f. { w : w不包含子字110 }

f. { w : w does not contain subword 110 }

g. { w : w的长度最多为 5}

g. { w : length of w is at most 5}

h. { w : w是除11和111之外的任何单词}

h. { w : w is any word except 11 and 111 }

i. { w : w 中的每个奇数位置都是1 }

i. { w : every odd position in w is 1 }

j. { w : w至少包含两个0，最多一个1 }

j. { w : w contains at least two 0s and at most one 1 }

k. { ε , 0 }

k. { ε , 0 }

l. { w : w包含偶数0或恰好包含两个1 }

l. { w : w contains an even number of 0s or contains exactly two 1s }

米 。

m. Empty set

名词​ 除空白外的所有单词

n . All words except empty

1.7 提供 NFA 的状态图，以识别包含给定数量状态的每种语言。在所有点上，字母表都是 { 0 , 1 }。

1.7 Give state diagrams for NFAs recognizing each of the languages ​​shown, containing the given number of states. At all points, the alphabet is { 0 , 1 }.

^A a. 语言 { w : w以00结尾} 具有三种状态

^A a. Language { w : w ends with 00 } with three states

b. 练习 1.6 中的语言有五种状态

b. The language from exercise 1.6c with five states

c. 练习 1.6l 中的语言有六种状态

c. The language from exercise 1.6l with six states

d.具有两种状态的 语言{0}

d. Language { 0 } with two states

e. 语言0 ^* 1 ^* 0 ⁺具有三种状态

e. Language 0 ^* 1 ^* 0 ⁺ with three states

^A f. 语言1 ^* ( 001 ⁺ ) ^*具有三种状态

^A f. Language 1 ^* ( 001 ⁺ ) ^* with three states

g.具有一种状态的 { ε } 语言

g. Language { ε } with one state

h. 语言0 ^*具有一种状态

h. Language 0 ^* with one state

1.8 使用定理 1.45 的证明中所示的结构，提供识别 中描述的语言之和的 NFA 的状态图

1.8 Use the construction given in the proof of Theorem 1.45 to give state diagrams of NFAs recognizing the union of languages ​​described in

练习 1.6a 和 1.6b。

a. exercises 1.6a and 1.6b.

b. 练习 1.6c 和 1.6f。

b. exercises 1.6c and 1.6f.

1.9 使用定理 1.47 的证明中所示的结构提供识别中描述的语言串联的 NFA 状态图

1.9 Use the construction shown in the proof of Theorem 1.47 to give state diagrams of NFAs that recognize concatenation of the languages ​​described in

a. 练习 1.6g 和 1.6i。

a. exercises 1.6g and 1.6i.

b. 练习 1.6b 和 1.6m。

b. exercises 1.6b and 1.6m.

1.10 使用定理 1.49 证明中所示的结构提供识别 Kleene 闭包中描述的语言的NFA 状态图

1.10 Use the construction shown in the proof of Theorem 1.49 to give state diagrams of NFAs recognizing Kleene closure of the languages ​​described in

练习 1.6b。

a. exercise 1.6b.

练习1.6j 。

b. exercise 1.6j.

c. 练习1.6m。

c. exercise 1.6m.

^A 1.11 证明任何NFA都可以转换为仅包含一个接受状态的等效非确定性有限状态机。

^A 1.11 Prove that every NFA can be converted into an equivalent nondeterministic finite automaton containing only one accepting state.

1.12 令D = { w : w包含偶数a和奇数b并且不包含子字ab }。提供识别语言D 的五态 DFA 和生成D的正则表达式。 （建议：尝试用更简单的方式描述D语言。）

1.12 Let D = { w : w contains an even number of a's and an odd number of b's and does not contain a substring ab }. Give a five-state DFA that recognizes the language D and a regular expression that generates D . (Suggestion: try to describe the language D in a simpler way.)

1.13 设F为字母表 { 0 , 1 } 上所有单词的语言，使得它们不包含由奇数个符号分隔的成对单词。提供识别F的五状态 DFA 的状态图。 （首先找到F 的补码的四状态 NFA 可能会有所帮助。）

1.13 Let F be the language of all words over the alphabet { 0 , 1 } such that they contain no pairs of ones separated by an odd number of symbols. Give a state diagram of a five-state DFA that recognizes F . (It may be helpful to first find a four-state NFA for the complement of F .)

1.14 a. 证明如果M是识别语言B 的DFA ，则交换M的接受和不接受状态给出一个识别语言B的补语的新 DFA 。由此推论，正则语言的类对于补码是闭集的。

1.14 a. Show that if M is a DFA recognizing language B , interchanging the accepting and nonaccepting states of M gives a new DFA recognizing the complement of language B . Conclude from this that the class of regular languages ​​is closed under complement.

b. 通过例子证明，如果M是一个识别语言C的 NFA ，则交换接受和不接受状态不一定会产生一个识别语言C的补集的新 NFA。 NFA认可的语言类别是否因互补而封闭？解释一下答案。

(b) Show by example that if M is an NFA that recognizes the language C , swapping the accepting and nonaccepting states does not necessarily produce a new NFA that recognizes the complement of C . Is the class of languages ​​recognized by an NFA closed under complement? Explain your answer.

1.15 给出一个反例，说明下面的构造不能证明定理1.49，即正则语言的类由于星号操作而封闭⁷。令N ₁ = ( Q ₁ , Σ , δ ₁ , q ₁ , F ₁ ) 识别语言A ₁。我们构造N = ( Q ₁ , Σ , δ , q ₁ , F ) 如下。N应该识别语言A ^* ₁。

1.15 Give a counterexample to show that the following construction does not allow us to prove Theorem 1.49 that the class of regular languages ​​is closed under the asterisk operation ⁷ . Let N ₁ = ( Q ₁ , Σ , δ ₁ , q ₁ , F ₁ ) recognize the language A ₁ . We construct N = ( Q ₁ , Σ , δ , q ₁ , F ) as follows. N is to recognize the language A ^* ₁ .

a. N 个 状态是N _{1 个}状态。

a. The N states are N ₁ states .

b. 初始状态N与初始状态N ₁相同。

b. The initial state of N is the same as the initial state of N ₁ .

c. F = { q ₁ } ∪ F ₁。

c. F = { q ₁ } ∪ F ₁ .

接受状态集合F由旧接受状态和初始状态组成。

The set of accepting states F consists of the old accepting states and the initial state.

d. 我们定义δ，使得对于任何q ∈ Q ₁和任何a ∈ Σ _ε，

d. We define δ so that for any q ∈ Q ₁ and any a ∈ Σ _ε ,

（建议：以图形方式显示此结构，如图 1.50 所示。）

(Suggestion: Show this construction graphically, as in Figure 1.50.)

1.16 使用定理 1.39 中给出的构造将以下两个非确定性有限自动机转换为等效的确定性有限自动机。

1.16 Use the construction in Theorem 1.39 to convert the following two nondeterministic finite-state machines into equivalent deterministic finite-state machines.

1.17 a.提供可识别语言 ( 01 ∪ 001 ∪ 010 ) ^* 的 NFA 。

1.17 a. Provide an NFA that recognizes the language ( 01 ∪ 001 ∪ 010 ) ^* .

b. 将此 NFA 转换为等效的 DFA。仅表示 DFA 中从初始状态可到达的部分。

b. Transform this NFA into an equivalent DFA. Show only the part of the DFA that is reachable from the initial state.

1.18 陈述生成练习1.6中语言的正则表达式。

1.18 Give the regular expressions that generate the languages ​​from Exercise 1.6.

1.19 使用引理 1.55 中描述的过程将以下正则表达式转换为非确定性有限自动机。

1.19 Use the procedure in Lemma 1.55 to transform the following regular expressions into nondeterministic finite automata.

( 0 ∪ 1 ) ^* 000 ( 0 ∪ 1 ) ^* ​

a. ( 0 ∪ 1 ) ^* 000 ( 0 ∪ 1 ) ^*

b. ((( 00 ) ^* ( 11 )) ∪ 01 ) ^*

b. ((( 00 ) ^* ( 11 )) ∪ 01 ) ^*

什么 ^*

What ^*​

1.20 对于以下每种语言，给出两个属于成员的单词和两个不是成员的单词 - 每点总共四个单词。假设字母表Σ  = { a , b } 在所有点上。

1.20 For each of the following languages, give two words that are members of it and two that are not members of it – four words in total for each point. Assume that the alphabet Σ  = { a , b } at all points.

a.a ^* b ^*

a. a ^* b ^*

b. a( ba ) ^* b。

b. a( ba ) ^* b

c. a ^* ∪ b ^*

c. a ^* ∪ b ^*

d. （aaa）^*

d. ( aaa ) ^*

e. Σ ^* a Σ ^* b Σ ^* a Σ ^*

e. Σ ^* a Σ ^* b Σ ^* a Σ ^*

f . 巴∪ 巴布

f. a ba ∪ bab

g. ( ε ∪ a ) b。

g. ( ε ∪ a ) b

h. ( a ∪ ba ∪ bb ) Σ ^*

h. ( a ∪ ba ∪ bb ) Σ ^*

1.21 使用引理 1.60 中描述的过程将以下有限自动机转换为正则表达式。

1.21 Use the procedure in Lemma 1.60 to transform the following finite automata into regular expressions.

1.22 在某些编程语言中，注释出现在分隔符之间，例如/#和#/。令C为所有有效限制注释词的语言。成员C必须以/#开头并以#/结尾，但其中不包含子词#/。为简单起见，假设C的字母表为Σ = { a , b , / , # }。

1.22 In some programming languages, comments appear between delimiters such as /# and #/ . Let C be the language of all valid delimited comment words. A member of C must begin with /# and end with #/ , but without an intervening subword #/ . For simplicity, assume that the alphabet of C is Σ = { a , b , / , # }.

a.提供可识别C 的 DFA 。

a. Provide a DFA that recognizes C .

b.指定生成C 的正则表达式。

b. Enter a regular expression that generates C .

^A 1.23 设B为字母表Σ上的任何语言。证明B = B ⁺当且仅当BB ⊆ B。

^A 1.23 Let B be an arbitrary language over the alphabet Σ . Prove that B = B ⁺ if and only if BB ⊆ B .

1.24 有限状态转换器（ FST ）是一种确定性有限自动机，其输出是某个单词，而不仅仅是接受或拒绝输入单词。T ₁和T ₂有限换能器的状态图如下所示。

1.24 A finite state transducer ( FST) is a type of deterministic finite state machine whose output is a word rather than merely accepting or rejecting an input word. State diagrams of finite state transducers T ₁ and T ₂ are shown below .

FST 中的每个转换由两个符号描述，其中一个表示该转换的输入符号，另一个表示输出符号。这两个符号由斜杠 / 分隔。在T _{1 处，从状态}q ₁到q ₂的转变具有输入符号2和输出符号1。一些转换可能包含多个输入输出对，例如，机器T _1中从q ₁到其自身的转换。当FST处理输入字w时，它一一读取输入符号w ₁ ... w _n，并从初始状态开始，执行与读取的符号w ₁ ... w _n = w匹配的转换。对于每个转换，它都会打印与该转换相对应的输出符号。例如，对于输入2212011，机器T ₁经历状态序列q ₁ , q ₂ , q ₂ , q ₂ , q ₂ , q ₁ , q ₁ , q ₁并产生输出1111000。机器T ₂对于输入abbb返回1011。为以下每个点提供状态字符串和返回的输出。

Each transition in the FST is described by two symbols, one of which designates the input symbol for the transition, and the other of which designates the output symbol. The two symbols are separated by a slash character, /. In T _{1 ,} the transition from state q ₁ to q ₂ has input symbol 2 and output symbol 1 . Some transitions may contain multiple input-output pairs, e.g., the transition from q ₁ to itself in machine T ₁ . When the FST processes the input word w , it reads the input symbols w ₁ … w _n in turn and, starting from the initial state, performs the transitions that match the symbols read w ₁ … w _n = w . For each transition, it prints out the output symbol corresponding to the transition. For example, for input 2212011 , _machine T 1 _traverses the sequence of states q ₁ , q ₂ , q ₂ , q ₂ , q ₁ , q ₁ , q 1 and produces the output 1111000 _. Machine T ₂ for input abbb returns 1011 . Give the sequence of states and the returned output for each of the following points.

a. T ₁用于输入011

a. T ₁ for input 011

b. T ₁用于输入211

b. T ₁ for input 211

c. T ₁用于输入121

c. T ₁ for input 121

d. T ₁用于输入0202

d. T ₁ for input 0202

e. T ₂用于输入b

e. T ₂ for input b

f. T ₂输入bbab

f. T ₂ for input bbab

g. T ₂输入bbbbbb

g. T ₂ for input bbbbbb

h. 输入ε的T ₂

h. T ₂ for input ε

1.25 阅读练习 1.24 中有限传感器的非正式定义。按照定义 1.5（第 35 页）中的模式提供此模型的正式定义。假设 FST 有一个输入字母表Σ和一个输出字母表Γ，但没有一组接受状态。包括 FST 执行的计算的正式定义。 （提示：FST 是有序五。其传递函数的形式为δ : Q × Σ → Q × Γ。）

1.25 Read the informal definition of a finite-state transducer in Exercise 1.24. Give a formal definition of this model using the pattern in Definition 1.5 (page 35). Assume that the FST has an input alphabet Σ and an output alphabet Γ , but no set of accepting states. Consider the formal definition of the computations performed by the FST. (Hint: The FST is an ordered quintuplet. Its transfer function is of the form δ : Q × Σ → Q × Γ . )

1.26使用练习 1.25 的解决方案，对练习 1.24 中所示的 机器T ₁和T _{2进行正式描述。}

1.26 Using the solution to Exercise 1.25, give a formal description of the machines T ₁ and T ₂ illustrated in Exercise 1.24.

1.27 阅读练习 1.24 中有限传感器的非正式定义。提供具有以下行为的 FST 状态图。它的输入和输出字母表是{ 0 , 1 }。其输出字在偶数位置与输入字相同，但在奇数位置包含反向符号。例如，对于输入0000111，此 FST 应返回1010010。

1.27 Read the informal definition of a finite-state transducer in Exercise 1.24. Give a state diagram of an FST with the following behavior. Its input and output alphabets are { 0 , 1 }. Its output word is identical to the input in even positions but contains inverse symbols in odd positions. For example, for an input of 0000111, this FST should return 1010010 .

1.28 使用定理 1.54 中描述的过程将以下正则表达式转换为 NFA。在所有点Σ = { a , b }。

1.28 Convert the following regular expressions to NFAs using the procedure in Theorem 1.54. At all points Σ = { a , b }.

a. a( abb ) ^* ∪ b

a. a( abb ) ^* ∪ b

b. a ⁺ ∪ ( ab ) ⁺

b. a ⁺ ∪ ( ab ) ⁺

( a ∪ b ⁺ ) a ⁺ b ⁺ ​

c. ( a ∪ b ⁺ ) a ⁺ b ⁺

1.29 使用泵引理证明下列语言不是正则语言。

1.29 Use the pumping lemma to show that the following languages ​​are not regular.

^A a 1 ₌ { 0 ⁿ 1 ⁿ 2 ⁿ : n ≥ 0}

^A a. A ₁ = { 0 ⁿ 1 ⁿ 2 ⁿ : n ≥ 0}

b. A ₂ = { www : w ∈ { a , b } ^* }

b. A ₂ = { www : w ∈ { a , b } ^* }

^A c. A ₃ = { a ^{2 n} : n ≥ 0} （这里a ^{2 n}表示 2 ⁿ个字符a的单词。）

^A c. A ₃ = { a ^{2 n} : n ≥ 0} (Here a ^{2 n} means a word composed of 2n ^characters a . )

1.30 解释下列“证明”中的错误：0 ^* 1 ^*不是正则语言（这个错误一定存在，因为语言0 ^* 1 ^* 是正则语言）。证据是间接的。假设0 ^* 1 ^*是正则的。令p为0 ^* 1 ^*的泵浦长度，由泵浦引理得出。作为s，我们选择单词0 ^p 1 ^p。我们知道s是0 ^* 1 ^*的元素，但示例 1.73 表明s不能被泵浦。我们得到了一个矛盾，所以0 ^* 1 ^*不是正则语言。

1.30 Explain the error in the following "proof" that 0 ^* 1 ^* is not a regular language (such an error must exist, since the language 0 ^* 1 ^* is regular). Proof by contradiction. Assume that 0 ^* 1 ^* is regular. Let p be the pumping length of 0 ^* 1 ^* , which follows from the Pumping Lemma. We choose s as the word 0 ^p 1 ^p . We know that s is an element of 0 ^* 1 ^* , but Example 1.73 shows that s cannot be pumped. We get a contradiction, so 0 ^* 1 ^* is not a regular language.

任务

Tasks

1.31 对于任何单词w = w ₁ w ₂ …w _n w 的逆 ，写为w R，是包含逆序符号w _n …w ₂ w _{1 的}单词。对于任何语言A，令A R = { w R : w ∈ A }。证明如果A是正则，则A R也是正则。

1.31 For any word w = w ₁ w ₂ …w _{n ,} the inverse of w , written as w R , is the word containing the symbols in the reverse order, w _n …w ₂ w ₁ . For any language A, let A R = { w R : w ∈ A }. Show that if A is regular, then A R is also regular.

1.32 让

1.32 Let

Σ ₃包含大小为 3 的所有列向量，由 0 和 1 组成。带有Σ _{3 的}符号字给出三行 0 和 1。让我们假设每一行都是一个二进制数，并让

Σ ₃ contains all column vectors of size 3, consisting of zeros and ones. The symbol word of Σ ₃ gives three rows of zeros and ones. Assume that each row is a binary number and let

B = { w ∈ Σ 3 ^{* ：} w 的底行是顶两行的总和}。

B = { w ∈ Σ 3 ^* : the bottom row of w is the sum of the top two rows }.

例如

For example

证明B是正则语言。 （提示：使用 BR 语言会更容易；您可以假设问题 1.31 中声明的结果。）

Prove that B is a regular language. (Hint: it will be easier to work with the language B R ; you can assume the result declared in Exercise 1.31.)

1.33 让

1.33 Let

在此示例中，Σ ₂包含高度为 2 的所有零和一的列向量。Σ ₂的符号字产生两行零和一。让我们假设每一行都是一个二进制数，并让

In this example, Σ ₂ contains all column vectors of zeros and ones of height two. The symbol word of Σ ₂ forms two rows of zeros and ones. Assume that each row is a binary number and let

C = { w ∈ Σ ₂ ^{* :} w 的底行是顶行的三倍}。

C = { w ∈ Σ ₂ ^* : the bottom row of w is triple the top row}.

例如

For example

但

证明C语言是正则的。 （你可以接受问题1.31中声明的结果作为已证明的结果。）

1.34 设Σ ₂与问题 1.33 中的字母表相同。让我们假设每一行都是一个二进制数，并让

1.34 Let Σ ₂ be the same alphabet as in Problem 1.33. Assume that each row is a binary number and let

D = { w ∈ Σ ₂ ^*：顶行是大于底行的数字}。

D = { w ∈ Σ ₂ ^* : top row is a number greater than bottom row}.

例如

For example

但

证明D是正则语言。

1.35 设Σ ₂与问题 1.33 中的字母表相同。假设顶行和底行是由 0 和 1 组成的单词，然后让

1.35 Let Σ ₂ be the same alphabet as in Problem 1.33. Assume that the top and bottom rows are words of zeros and ones, and let

E = { w ∈ Σ ₂ ^{* :} w的底线是w顶线的倒数}。

E = { w ∈ Σ ₂ ^* : the bottom row of w is the inverse of the top row of w }.

证明语言E不是正则语言。

Show that the language E is not regular.

1.36 令B _n = {且^k：k是n 的倍数}。证明对于每个n ≥ 1，语言B _n是正则的。

1.36 Let B _n = { a ^k : k is a multiple of n }. Prove that for every n ≥ 1 the language B _n is regular.

1.37 设C _n = { x : x是一个二进制数，是n的倍数}。证明对于每个n ≥ 1，C语言_n都是正则的。

1.37 Let C _n = { x : x be a binary number that is a multiple of n }. Show that for every n ≥ 1 the language C _n is regular.

1.38 All-NFA M是有序五（Q , Σ , δ , q ₀ , F）。如果M在读取输入x后可能最终进入的每个可能状态都是属于F的状态，则M接受单词x ∈ Σ ^{*。与 All-NFA 不同，如果这些可能状态中}至少有一个是接受状态，则正则 NFA 会接受一个单词。证明 all-NFA 能够识别一类正则语言。

1.38 All-NFA M is an ordered 5 ( Q , Σ , δ , q ₀ , F ). M accepts a word x ∈ Σ ^* if every possible state that M can enter after reading input x is a state in F . In contrast to All-NFA, an ordinary NFA accepts a word if at least one of these possible states is an accepting state. Prove that All-NFA recognizes the class of regular languages.

1.39 定理 1.54 所示的结构表明，每个 GNFA 等价于仅包含两个状态的 GNFA。我们可以证明 DFA 存在相反的现象。证明对于每个k > 1，都存在一种语言A _k ⊆ { 0 , 1 } ^*，使得它可以被包含k 个状态的 DFA 识别，但不能被仅包含k − 1 个状态的 DFA 识别。

1.39 The construction in Theorem 1.54 shows that every GNFA is equivalent to a GNFA with only two states. We can show that the opposite holds for DFAs. Prove that for every k > 1 there exists a language A _k ⊆ { 0 , 1 } ^* such that it is recognized by a DFA with k states but not by one with only k − 1 states.

1.40 回想一下，如果存在一个单词z使得xz = y，则x是单词y的前缀，并且如果另外x ≠ y ，则x是y 的真前缀。在以下每一点中，我们定义了语言A上的某种操作。证明正则语言的类因该操作而封闭。

1.40 Recall that x is a prefix of a word y if there is a word z such that xz = y , and x is a proper prefix of y if, in addition, x ≠ y . In each of the following steps we define some operation on the language A . Show that the class of regular languages ​​is closed under this operation.

^A a. NOPREFIX( A ) = { w ∈ A : 没有特定的前缀w属于语言A }。

^A a.NOPREFIX ( A ) = { w ∈ A : no proper prefix w belongs to language A }.

b. NOEXTEND( A ) = { w ∈ A : w不是属于A 的任何单词的正确前缀}。

b. NOEXTEND( A ) = { w ∈ A : w is not a proper prefix of any word belonging to A }.

1.41 我们将语言A和 B的完美重新洗牌称为语言

1.41 We will call the perfect shuffle of languages ​​A and  B the language

{ w : w = a ₁ b ₁ …a _k b _k，其中a ₁ … a _k ∈ A且b ₁ … b _k ∈ B对于每个a _i , b _i ∈ Σ }。

{ w : w = a ₁ b ₁ …a _k b _k , where a ₁ … a _k ∈ A and b ₁ … b _k ∈ B for every a _i , b _i ∈ Σ }.

证明正则语言的类在完美洗牌下是封闭的。

Prove that the class of regular languages ​​is closed under perfect shuffling.

1.42 语言A和 B的重新洗牌称为语言

1.42 The reshuffling of languages ​​A and  B is called the language

{ w : w = a ₁ b ₁ …a _k b _k，其中a ₁ … a _k ∈ A且b ₁ … b _k ∈ B对于每个a _i , b _i ∈ Σ ^* }。

{ w : w = a ₁ b ₁ …a _k b _k , where a ₁ … a _k ∈ A and b ₁ … b _k ∈ B for every a _i , b _i ∈ Σ ^* }.

证明正则语言的类由于重新洗牌而关闭。

Prove that the class of regular languages ​​is closed under reshuffling.

1.43 设A为任意语言。我们将 DROP-OUT( A ) 定义为一种包含所有单词的语言，这些单词可以通过从属于A 的单词中删除一个符号来获得。因此 DROP-OUT( A ) = { xz : xyz ∈ A，其中x , z ∈ Σ ^* , y ∈ Σ }。证明正则语言的类由于DROP-OUT操作而封闭。给出一个图形证明和一个更正式的构造证明，如定理 1.47 中所示。

1.43 Let A be an arbitrary language. We define DROP-OUT( A ) to be the language containing all words that can be obtained by removing one symbol from a word in A . So DROP-OUT( A ) = { xz : xyz ∈ A , where x , z ∈ Σ ^* , y ∈ Σ }. Prove that the class of regular languages ​​is closed under the DROP-OUT operation. Give both a proof by drawing and a more formal proof by construction, as in Theorem 1.47.

^A 1.44 设B和C是字母表上的语言Σ = { 0 , 1 }。我们定义

^A 1.44 Let B and C be languages ​​over the alphabet Σ = { 0 , 1 }. We define

B ← ¹ C = { w ∈ B：对于某些y ∈ C，单词w和 y包含相同数量的 1}。

B ← ¹ C = { w ∈ B : for some y ∈ C the words w and  y contain equal numbers of ones}.

证明正则语言的类由于操作¬ ¹而封闭。

Prove that the class of regular languages ​​is closed under the operation ¬ ¹ .

*1.45 设A/B = { w : wx ∈ A对于某些x ∈ B }。证明如果A是正则语言，B是任意语言，则A/B是正则语言。

*1.45 Let A/B = { w : wx ∈ A for some x ∈ B }. Prove that if A is a regular language and B is any language, then A/B is a regular language.

1.46 证明下列语言不是正则语言。您可以使用泵引理和正则语言类的闭包来实现并集、交集和补集。

1.46 Prove that the following languages ​​are not regular. You can use the pumping lemma and the closure of the class of regular languages ​​under union, intersection, and complement.

{ 0 n 1 ^m 0 ⁿ : m , n ^≥ 0}

a. { 0 ⁿ 1 ^m 0 ⁿ : m , n ≥ 0}

{ 0 ^m 1 ⁿ : ^m ≠ n }​​

^A b. { 0 ^m 1 ⁿ : m ≠ n }

c. { w : w ∈ { 0 , 1 } ^*不是回文} ⁸

c. { w : w ∈ { 0 , 1 } ^* is not a palindrome} ⁸

* d。 { wtw : w , t ∈ { 0 , 1 } ⁺ }

* d . { wtw : w , t ∈ { 0 , 1 } ⁺ }

1.47 令Σ = { 1 , # } 并令

1.47 Let Σ = { 1 , # } and let

Y = { w : w = x ₁ # x ₂ # … # x _k对于k ≥ 0，每个x _i ∈ 1 ^*且x _i ≠ x _j对于i ≠ j }。

Y = { w : w = x ₁ # x ₂ # … # x _k for k ≥ 0, each x _i ∈ 1 ^* and x _i ≠ x _j for i ≠ j }.

证明Y不是正则语言。

Prove that Y is not a regular language.

1.48 令Σ = { 0 , 1 } 并令

1.48 Let Σ = { 0 , 1 } and let

D = { w : w包含相同次数的子字01和10 }。

D = { w : w contains an equal number of occurrences of the substrings 01 and 10 }.

这意味着101 ∈ D，因为101包含 1 次出现的01和 1 次10，但1010 ∉ D，因为1010包含两次出现的10和 1 次01。证明D是正则语言。

This means that 101 ∈ D , since 101 contains one occurrence of 01 and one occurrence of 10 , but 1010 ∉ D , since 1010 contains two occurrences of 10 and one occurrence of 01 . Show that D is a regular language.

1.49 a. 设B = { 1 ^k y : y ∈ { 0 , 1 } ^*且y至少包含k个 1，k ≥ 1}。证明B是正则语言。

1.49 a. Let B = { 1 ^k y : y ∈ { 0 , 1 } ^* and y contains at least k 1s, k ≥ 1}. Prove that B is a regular language.

b. 设B = { 1 ^k y : y ∈ { 0 , 1 } ^*且y最多包含k个，k ≥ 1}。证明B不是正则语言。

b. Let B = { 1 ^k y : y ∈ { 0 , 1 } ^* and y contains at most k 1s, k ≥ 1}. Prove that B is not a regular language.

^A 1.50 阅读练习 1.24 中给出的有限传感器的非正式定义。如果输入和输出字母表为 { 0 , 1 }，则 FST 不能为每个输入单词w返回w R 。

^A 1.50 Read the informal definition of a finite-state transducer given in Exercise 1.24. Prove that no FST can return w for each input word w if the input and output alphabets are { 0 , 1 }.

1.51 设x和y为单词，L为任意语言。如果存在单词z并且单词xz和yz中的一个恰好属于L，则我们说x和y可通过 L 区分；否则，如果对于每个单词z我们得到xz ∈ L当且仅当yz ∈ L时，那么我们说x和y无法通过L区分。如果x和y无法通过L区分，我们将其写为x ≡ _L y。证明≡ _L是等价关系。

1.51 Let x and y be words and L an arbitrary language. We say that x and y are distinguishable by L if there is a word z such that exactly one of the words xz and yz is in L ; otherwise, if for every word z we get xz ∈ L if and only if yz ∈ L , then we say that x and y are indistinguishable by L. If x and y are indistinguishable by L , we write this as x ≡ _L y . Show that ≡ _L is an equivalence relation.

^A* 1.52 迈希尔-尼罗德定理。返回任务 1.51。设L是一种语言， X是一组单词。如果属于X的每两个不同单词都可以通过L区分则我们说X可以通过L成对区分。让我们将索引 L定义为可通过L成对区分的任何集合的最大元素数。 L索引可以是有限的或无限的。

^A* 1.52 Myhill–Nerode Theorem . Return to Problem 1.51. Let L be a language and X be a set of words. We say that X is pairwise distinguishable by L if every two distinct words belonging to X are distinguishable by L . Let us define the index L as the maximum number of elements of any set that is pairwise distinguishable by L . The index L may be finite or infinite.

a. 证明如果L被k 个状态的 DFA 识别，则L的索引至多为k。

a. Prove that if L is recognized by a k- state DFA, then the index of L is at most k .

b. 证明如果索引L是有限数k ，则它可以被包含k 个状态的 DFA 识别。

b. Prove that if the index L is a finite number k , then it is recognized by a DFA containing k states.

c. 得出L是正则的当且仅当它具有有限索引时。而且，这个索引是识别这种语言的最小DFA的大小。

c. Deduce that L is regular if and only if it has a finite index. Moreover, this index is the size of the smallest DFA that recognizes this language.

1.53 设Σ = { 0,1 , + , = } i

1.53 Let Σ = { 0 , 1 , + , = } and

ADD = { x = y + z : x , y , z是二进制整数，x是y和z的和}。

ADD = { x = y + z : x , y , z are binary integers and x is the sum of y and z }.

表明ADD不是正则语言。

Show that ADD is not a regular language.

1.54 考虑语言F = { a ⁱ b ^j c ^k : i , j , k ≥ 0 且如果i = 1，则j = k }。

1.54 Consider the language F = { a ⁱ b ^j c ^k : i , j , k ≥ 0 and if i = 1, then j = k }.

a. 证明F不是正则语言。

a. Show that F is not a regular language.

b. 证明对于泵引理，语言 F 的行为就像它是正则的一样。换句话说，说明泵浦长度并证明F满足该p值的泵浦引理的三个条件。

(b) Show that with respect to the Pumping Lemma, the language F behaves as if it were regular. In other words, give the pumping length and demonstrate that F satisfies the three conditions of the Pumping Lemma for this value of p .

c. 解释为什么(a)和(b)与泵引理不矛盾。

c. Explain why points (a) and (b) do not contradict the pumping lemma.

1.55 泵引理指出，每种正则语言都有泵长度p，这样该语言中的任何单词如果长度为p或更大，则可以被泵泵。如果p是语言A的泵浦长度，则任何长度p' ≥ p也是如此。语言A的最小泵浦长度是最小的p，即A的泵浦长度。例如，如果A = 01 ^*，则最小膨胀长度为 2。原因是单词s = 0属于A，长度为 1，但s不能膨胀；而A中长度为 2 或以上的每个单词都包含 1，因此可以通过将其除以x = 0、y = 1和z来泵送该单词的余数。提供以下每种语言的最小泵送长度，并解释您的答案。

1.55 The pumping lemma states that every regular language has a pumping length p such that every word of the language can be pumped if it has length p or greater. If p is a pumping length for language A , then so is any length p' ≥ p . The minimal pumping length for language A is the smallest p that is a pumping length of A . If, for example, A = 01 ^* , the minimal pumping length is 2. The reason is that the word s = 0 belongs to A and has length 1, but s cannot be pumped; on the other hand, every word in A of length 2 or greater contains a 1 and thus can be pumped by partitioning it so that x = 0 , y = 1 , and z is the remainder of the word. Give the minimal pumping length for each of the following languages, giving an explanation of your answer.

^{0001 *}​ ​

^A a. 0001 ^*

0 ^{* 1 *}​​

^A b. 0 ^* 1 ^*

c. 001 ∪ 0 ^* 1 ^*

c. 001 ∪ 0 ^* 1 ^*

^且0 ^* 1 ⁺ 0 ⁺ 1 ^* ∪ 10 ^* 1 。

^And d. 0 ^* 1 ⁺ 0 ⁺ 1 ^* ∪ 10 ^* 1

e. ( 01 ) ^*

e. ( 01 ) ^*

f 。

f. ε

1 ^* 01 ^* 01 ^* ​

g. 1 ^* 01 ^* 01 ^*

10 ( 11 ^* 0 ) ^* 0​

h. 10 ( 11 ^* 0 ) ^* 0

一、 1011

and 1011

j ^。​

j. Σ ^*

^* 1.56 如果A是一些自然数的集合， k是大于 1 的自然数，则令

^* 1.56 If A is a set of some natural numbers and k is a natural number greater than 1, let

B _k ( A ) = { w : w是A中某个数以k为基数的表示}。

B _k ( A ) = { w : w is the base k representation of some number in A }.

在这种情况下，我们不允许在数字表示中使用前导零。例如，B ₂ ({ 3 , 5 }) = { 11 , 101 } 且B ₃ ({ 3 , 5 }) = { 10 , 12 }。给出一个集合A的示例，其中B ₂ ( A ) 是正则语言，但B ₃ ( A ) 不是正则语言。证明例子的正确性。

In this case we do not allow leading zeros in the representation of the number. For example, B ₂ ({ 3 , 5 }) = { 11 , 101 }, and B ₃ ({ 3 , 5 }) = { 10 , 12 }. Give an example of a set A for which B ₂ ( A ) is regular but B ₃ ( A ) is not regular. Prove the correctness of the example.

^* 1.57 设A为任何语言；令A A Σ为属于A 的所有前半字的集合，使得

^* 1.57 Let A be any language; let A A Σ be the set of all first half-words belonging to A , such that

A A = { x : 对于某些y , | x | = | y |且xy ∈ A }。

A A = { x : for some y , | x | = | y | and xy ∈ A }.

证明如果A是正则，则A A也是正则。

Prove that if A is regular, then A A is also regular.

^* 1.58 设A为任何语言；设A B−B为A中所有单词的集合，去掉中间三分之一，即

^* 1.58 Let A be any language; let A B−B be the set of all words in A with the middle third removed, that is,

A B−B = { xz：对于某些y， | x | = | y | = |与|且xyz ∈ A }。

A B−B = { xz : for some y , | x | = | y | = | z | and xyz ∈ A }.

证明如果A是正则语言，则A B−B不必是正则语言。

Prove that if A is a regular language, A B−B need not be regular.

^* 1.59 设M = ( Q , Σ , δ , q ₀ , F ) 为 DFA， h为M的状态，称为“home”。 M和h的同步序列是单词s ∈ Σ ^*，其中对于每个q ∈ Q δ ( q , s ) = h 。 [这里我们将函数δ的定义扩展到单词中，使得δ ( q , s ) 等于M从状态q开始并读取输入单词s时结束的状态。] 我们说M是可同步的如果它包含某个状态h的同步字符串。证明如果M是一个k态可同步DFA，那么它有一个长度至多为k ³的同步序列。这个估计可以改进吗？

^* 1.59 Let M = ( Q , Σ , δ , q ₀ , F ) be a DFA and let h be a state of M called the "home" state. The synchronizing sequence for M and h is a word s ∈ Σ ^* , where δ ( q , s ) = h for all q ∈ Q . [Here we extend the definition of the function δ to words, so that δ ( q , s ) is equal to the state that M ends up in when it starts in state q and reads input word s .] We say that M is synchronizable if it contains a synchronizing sequence for some state h . Prove that if M is a k -state synchronizable DFA, then it has a synchronizing sequence of length at most k ³ . Can this estimate be improved?

1.60 设Σ = { a , b }。对于每个k ≥ 1，令C _k是由所有在从右端算起的第 k 个位置包含a 的单词组成的语言。换句话说，C _k = Σ ^* a Σ ^{k -1}。描述一个识别 C 语言的k +1 状态 NFA _，既作为状态图又作为形式描述。

1.60 Let Σ = { a , b }. For all k ≥ 1, let C _k be the language of all words containing a in exactly the kth position from the right end. In other words, C _k = Σ ^* a Σ ^{k −1} . Describe an NFA with k + 1 states that recognizes the language C _k , both as a state diagram and as a formal description.

1.61 考虑问题 1.60 中定义的C _k语言。证明对于每个k，状态少于 2 ^{k 的}DFA无法识别C _k。

1.61 Consider the languages ​​C _k defined in Problem 1.60. Prove that for each k no DFA with fewer than 2 ^k states can recognize C _k .

1.62 令Σ = { a , b }。对于任何k ≥ 1，令D _k是一种由最后k 个符号中至少包含一个a 的所有单词组成的语言。换句话说，D _k = Σ ^* a ( Σ ∪ ε ) ^{k -1}。描述一个最多包含k +1 个状态的 DFA，该状态识别D _k，既作为状态图又作为形式描述。

1.62 Let Σ = { a , b }. For each k ≥ 1, let D _k be the language of all words that contain at least one a among the last k symbols. In other words, D _k = Σ ^* a ( Σ ∪ ε ) ^{k −1} . Describe a DFA with at most k +1 states that recognizes D _k as both a state diagram and a formal description.

^* 1.63 a. 设A为无限正则语言。证明A可以分为两个无限的、不相交的正则子集。

^* 1.63 a. Let A be an infinite regular language. Prove that A can be divided into two infinite, disjoint regular subsets.

b. 设B和D是两种语言。如果B ⊆ D并且D包含无限多个不属于B 的单词，我们将写成B ⋐ D。证明如果B和D是两种正则语言且B ⋐ D，我们可以找到正则语言C使得B ⋐ C ⋐ D。

b. Let B and D be two languages. We write B ⋐ D if B ⊆ D and D contains infinitely many words that are not members of B . Show that if B and D are two regular languages, where B ⋐ D , we can find a regular language C such that B ⋐ C ⋐ D .

1.64 令N为包含k 个状态的 NFA ，其识别某种语言A。

1.64 Let N be an NFA containing k states that recognizes some language A .

a. 证明如果A非空，则A最多包含一个长度为k的单词。

a. Prove that if A is nonempty, then A contains a word of length at most k .

*b。 通过示例表明，如果我们将A都替换为A ，则(a)点不一定为真。

*b. Show with an example that point (a) does not have to be true if we replace both A in it with A .

c. 证明如果A非空，则A最多包含一个长度为^2k的单词。

c. Prove that if A is nonempty, then A contains a word of length at most ^2k .

*d。表明(c) 中给出的限制几乎是严格的；换句话说，对于每个k，显示一个识别语言A _{k 的}NFA ，其中A k是非空的，并且属于A k 的最短单词的长度相对于k呈指数增长。尝试尽可能接近(c)中的极限。

*d. Show that the constraint given in (c) is almost strict; in other words, for each k show an NFA that recognizes a language A _k , where A k is nonempty and the shortest words of A k have length exponential in k . Try to get as close to the constraint in (c) as possible.

^* 1.65 证明对于每个n > 0 都存在一种语言B _n使得

^* 1.65 Prove that for every n > 0 there exists a language B _n such that

a. B _n由包含n 个状态的 NFA 识别，

a. B _n is recognizable by an NFA containing n states,

b. 如果B _n = A ₁ ∪ … ∪ A _k，其中A_和是正则语言，则A_和中至少一个需要包含指数多个状态的 DFA。

b. If B _n = A ₁ ∪ … ∪ A _k , where A _and are regular languages, then at least one of these A _and requires a DFA containing exponentially many states.

1.66 同态是一个函数f : Σ → Γ ^*从一个字母表到另一个字母表上的一组单词。我们可以通过定义f ( w ) = f ( w ₁ ) f ( w ₂ )… f ( w _n ) 来扩展f对单词进行操作，其中w = w ₁ w ₂ …w _n且每个w _i ∈ Σ。我们可以通过为任何语言A定义f ( A ) = { f ( w ) : w ∈ A } 来进一步扩展f来处理语言。

1.66 A homomorphism is a function f : Σ → Γ ^* from one alphabet into the set of words over another alphabet. We can extend f to act on words by defining f ( w ) = f ( w ₁ ) f ( w ₂ )… f ( w _n ), where w = w ₁ w ₂ …w _n and each w _i ∈ Σ . We can further extend f to act on languages ​​by defining f ( A ) = { f ( w ) : w ∈ A }, for any language A .

a. 通过形式化构造证明正则语言的类由于同态而封闭。换句话说，给定一个识别B的 DFA M和一个同态f，构造一个识别f ( B )的有限自动机M' 。分析构造的机器M'。是否在所有情况下都是 DFA？

a. Show by formal construction that the class of regular languages ​​is closed under homomorphism. In other words, given a DFA M recognizing B and a homomorphism f , construct a finite automaton M' that recognizes f ( B ). Analyze the constructed automaton M' . Will it be a DFA in each case?

b. 举例说明非正则语言的类由于同态而不是封闭的。

b. Show by an example that the class of irregular languages ​​is not closed under homomorphism.

*1.67 令语言A的旋转闭包为RC ( A ) = { yx : xy ∈ A }。

*1.67 Let the rotational closure of A be RC ( A ) = { yx : xy ∈ A }.

a.证明对于任何语言 A RC ( A ) = RC ( RC ( A ) ) 成立。

a. Show that for any language A RC ( A ) = RC ( RC ( A )).

b. 证明正则语言的类由于旋转闭包而被封闭。

b. Show that the class of regular languages ​​is closed under rotational closure.

^* 1.68 在传统的洗牌方法中，一副牌被随机分成两部分，然后以相反的顺序折叠。在一种更高级的转移中，称为“斯卡恩转移”，甲板被分为三部分，当甲板组装时，中间的部分被放置在顶部。我们将使用斯卡恩的翻译作为一些舌头手术的灵感。对于语言A，令CUT ( A ) = { yxz : xyz ∈ A }。

^* 1.68 In the traditional method of cutting a deck of cards, it is randomly divided into two pieces, which are then reassembled in reverse order. In a more advanced cut, called the Scarne cut , the deck is divided into three pieces and the middle piece is placed on top when the deck is reassembled. We will use Scarne 's cut as inspiration for a certain operation on languages. For language A , let CUT ( A ) = { yxz : xyz ∈ A }.

a. 显示语言B，其中CUT ( B ) ≠ CUT ( CUT ( B ))。

a. Show language B for which CUT ( B ) ≠ CUT ( CUT ( B )).

b. 证明正则语言类在CUT下是封闭的。

b. Prove that the class of regular languages ​​is closed under CUT .

1.69 设Σ = { 0 , 1 }。设WW _k = { ww : w ∈ Σ ^*且w的长度为k }。

1.69 Let Σ = { 0 , 1 }. Let WW _k = { ww : w ∈ Σ ^* and w has length k }.

a. 证明对于任意k，状态数少于 2 ^{k 的}DFA无法识别WW _k。

a. Prove that for any k, no DFA with less than 2 ^k states can recognize WW _k .

b.描述WW _k 的更小的 NFA ，即WW _k的补集。

b. Describe a much smaller NFA for WW _k , the complement of WW _k .

1.70我们将语言A和 B的避免 操作定义为

1.70 We define the avoids operation for languages ​​A and  B as

A 避免 B = { w : w ∈ A并且 w不包含B的任何单词作为子词}。

A avoids B = { w : w ∈ A and  w does not contain any word from B as a subword}.

证明正则语言的类由于避免操作而封闭。

Prove that the class of regular languages ​​is closed under the avoids operation .

1.71 设Σ = { 0 , 1 }。

1.71 Let Σ = { 0 , 1 }.

a. 设A = { 0 ^k u 0 ^k : k ≥ 1 且u ∈ Σ ^* }。证明A是正则语言。

a. Let A = { 0 ^k u 0 ^k : k ≥ 1 and u ∈ Σ ^* }. Prove that A is a regular language.

b. 令B = { 0 ^k 1 u 0 ^k : k ≥ 1 且u ∈ Σ ^* }。证明B不是正则语言。

b. Let B = { 0 ^k 1 u 0 ^k : k ≥ 1 and u ∈ Σ ^* }. Prove that B is not a regular language.

1.72 令M ₁和M ₂分别为包含k ₁和k ₂状态的两个DFA，并令U = L ( M ₁ ) ∪ L ( M ₂ )。

1.72 Let M ₁ and M ₂ be two DFAs that contain k ₁ and k ₂ states, respectively, and let U = L ( M ₁ ) ∪ L ( M ₂ ).

a. 证明如果U ≠ Ø，则U包含序列s使得 | s | < 最大值 ( k ₁ , k ₂ )。

a. Prove that if U ≠ Ø , then U contains some sequence s such that | s | < max( k ₁ , k ₂ ).

b. 证明如果U ≠ Σ ^*，则U不包含某些单词s使得 | s | < k ₁ k ₂。

b. Prove that if U ≠ Σ ^* , then U does not contain some word s such that | s | < k ₁ k ₂ .

1.73 令Σ = { 0 , 1 , # }。令C = { x # x R # x| x Î { 0 , 1 }*}。证明C是 CFL

1.73 Let Σ = { 0 , 1 , # }. Let C = { x # x R # x| x Î { 0 , 1 }*}. Show that C is a CFL

精选解决方案

Selected solutions

1.1 对于M ₁：(a) q ₁；(b) { q ₂ }；(c) q ₁、q ₂、q ₃、q ₁、q ₁；(d)否；(e)否

1.1 For M ₁ : (a) q ₁ ; (b) { q ₂ }; (c) q ₁ , q ₂ , q ₃ , q ₁ , q ₁ ; (d) No; (e) No

对于M ₂：(a) q ₁；(b) { q ₁ , q ₄ }; (c) q ₁、q ₁、q ₁、q ₂、q ₄；(d)是；(e)否

For M ₂ : (a) q ₁ ; (b) { q ₁ , q ₄ }; (c) q ₁ , q ₁ , q ₁ , q ₂ , q ₄ ; (d) Yes; (e) No

1.2 M ₁ = ({ q ₁ , q ₂ , q ₃ }, { a , b }, δ ₁ , q ₁ , { q ₂ })。

1.2 M ₁ = ({ q ₁ , q ₂ , q ₃ }, { a , b }, δ ₁ , q ₁ , { q ₂ }).

M2 = ({ q ₁ , q ₂ , q ₃ , q ₄ }, { a , b }, δ ₂ , q ₁ , { q ₁ , q ₄ })。

M2 = ({ q ₁ , q ₂ , q ₃ , q ₄ }, { a , b }, δ ₂ , q ₁ , { q ₁ , q ₄ }).

过渡函数是

The transition functions are

δ1​ δ 1	和 and	乙 b		δ2​ δ 2	和 and	乙 b
q 1 q1​	q 2 q2​	q 1 q1​		q 1 q1​	q 1 q1​	q 2 q2​
q 2 q2​	q 3 q3​	q 3 q3​		q 2 q2​	q 3 q3​	q 4 q4​
q 3 q3​	q 2 q2​	q 1 q1​		q 3 q3​	q 2 q2​	q 1 q1​
				q 4 q4​	q 3 q3​	q 4 q4​

1.4 (b)下面显示的是两种语言的 DFA { w : w恰好包含两个a } 和 { w : w至少包含两个b }。

1.4 (b) Below is a DFA for two languages ​​{ w : w contains exactly two a's } and { w : w contains at least two b 's }.

使用交集结构将它们组合起来可得到以下 DFA。

Combining them using the intersection construction gives the DFA below.

尽管该任务不要求简化此 DFA，但可以组合某些状态以给出以下 DFA。

Although the problem does not require simplifying this DFA, certain states can be combined, resulting in the following DFA.

(d)下面显示的是两种语言的 DFA { w : w包含偶数个a } 和 { w : w中的 每个a后面至少有一个b }。

(d) Below are DFAs for two languages ​​{ w : w contains an even number of a } and { w : every a in  w is followed by at least one b }.

使用交集结构将它们组合起来可得到以下 DFA。

Combining them using the intersection construction gives the DFA below.

尽管该任务不要求简化此 DFA，但可以组合某些状态以给出以下 DFA。

Although the problem does not require simplifying this DFA, certain states can be combined, resulting in the following DFA.

1.5 (a)左侧所示的 DFA 识别语言 { w : w contains ab }。右 DFA 识别其补码 { w : w不包含ab }。

1.5 (a) The DFA shown on the left recognizes the language { w : w contains ab }. The right DFA recognizes its complement, { w : w does not contain ab }.

(b)以下 DFA 识别语言 { w : w contains baba }。

(b) The following DFA recognizes the language { w : w contains baba }.

该 DFA 识别语言 { w : w不包含baba }。

This DFA recognizes the language { w : w does not contain baba }.

1.7 (a)(f)

1.7 (a) (f)

1.11 令N = ( Q , Σ , δ , q ₀ , F ) 为任意 NFA。我们用单个接受状态构建 NFA N' ，该接受状态将识别与N相同的语言。非正式地，N'与N相同，除了它包含新的接受状态qaccept_{和从与接受}N的状态相对应的状态引导到新状态的ε-转换。 q_接受状态没有传出转换。更正式地说，N' = ( Q ∪ { q _Accept }, Σ , δ' , q ₀ , { q _Accept })，其中对于每个q ∈ Q和a ∈ Σ _ε

1.11 Let N = ( Q , Σ , δ , q ₀ , F ) be an arbitrary NFA. We construct an NFA N' with a single accepting state that recognizes the same language as N . Informally, N' is identical to N except that it contains a new accepting state q _accept and ε -transitions leading from states corresponding to the accepting states of N to the new state. The state q _accept has no outgoing transitions. More formally, N' = ( Q ∪ { q _accept }, Σ , δ' , q ₀ , { q _accept }), where for every q ∈ Q and a ∈ Σ _ε

且δ' ( q _Accept , a ) = Ø对于所有a ∈ Σ _ε。

and δ' ( q _accept , a ) = Ø for every a ∈ Σ _ε .

1.23 我们“当且仅当”证明两个方向。

1.23 We prove both directions "if and only if".

( ⇒ ) 我们假设B = B ⁺并证明BB ⊆ B。

( ⇒ ) We assume that B = B ⁺ and we will show that then BB ⊆ B .

对于每种语言BB ⊆ B ⁺都成立，因此如果B = B ⁺，则BB ⊆ B。

For every language, BB ⊆ B ⁺ holds , so if B = B ⁺ , then BB ⊆ B .

( Ü ) 我们假设BB ⊆ B并证明B = B ⁺。

( Ü ) We assume that BB ⊆ B and we will show that then B = B ⁺ .

因此，对于任何语言B ⊆ B ⁺，我们只需要证明B ⁺ ⊆ B。如果w ∈ B ⁺，则w = x ₁ x ₂ …x _k，其中每个x _i ∈ B且k ≥ 1。由于x ₁、x ₂ ∈ B且BB ⊆ B，我们得到x ₁ x ₂ ∈ B。类似地，由于x 1 _x 2_属于B且x ₃属于B ，所以我们得到x ₁ x ₂ x ₃ ∈ B。继续这样，我们得到x ₁ x ₂ …x _k ∈ B。因此w ∈ B，因此我们可以得出B ⁺ ⊆ B的结论。

For any language B ⊆ B ⁺ , therefore we need only show that B ⁺ ⊆ B . If w ∈ B ⁺ , then w = x ₁ x ₂ …x _k , where every x _i ∈ B and k ≥ 1. Since x ₁ , x ₂ ∈ B and BB ⊆ B , we get x ₁ x ₂ ∈ B . Analogously, since x ₁ x ₂ is a member of B and x ₃ is a member of B , we get x ₁ x ₂ x ₃ ∈ B . Continuing in this way, we get x ₁ x ₂ …x _k ∈ B . Hence w ∈ B , and we can conclude that B ⁺ ⊆ B .

第二个推理可以正式写为以下归纳证明。

This second reasoning can be written formally as the following proof by induction.

我们假设BB ⊆ B。

We assume that BB ⊆ B .

论文：对于每个k ≥ 1，如果x ₁ , …, x _k ∈ B，则x ₁ …x _k ∈ B。

Thesis : For every k ≥ 1, if x ₁ , …, x _k ∈ B , then x ₁ …x _k ∈ B .

基本步骤：证明k = 1。这个陈述显然是正确的。

Basic step : Proof for k = 1. This statement is obviously true.

归纳步骤：对于每个k ≥ 1，我们假设该命题对于k成立，并且我们证明它对于k + 1 成立。

Inductive step : For each k ≥ 1, we assume that the theorem is true for k , and prove that it is true for k + 1.

如果x ₁ , …, x _k , x _{k +1} ∈ B，则基于归纳假设x ₁ …x _k ∈ B。因此，x ₁ …x _k x _{k +1} ∈ BB，但BB ⊆ B，因此x ₁ …x _k x _{k +1} ∈ B。这样就完成了归纳步骤和论文的证明。该论文意味着如果BB ⊆ B，则B ⁺ ⊆ B。

If x ₁ , …, x _k , x _{k +1} ∈ B , then by the inductive assumption x ₁ …x _k ∈ B . Hence x ₁ …x _k x _{k +1} ∈ BB , but BB ⊆ B , therefore x ₁ …x _k x _{k +1} ∈ B . This completes the proof of the inductive step and the thesis. It follows from the thesis that if BB ⊆ B , then B ⁺ ⊆ B .

1.29 (a)我们假设语言A ₁ = { 0 ⁿ 1 ⁿ 2 ⁿ : n ≥ 0} 是正则语言。令p为由泵浦引理得出的泵浦长度。我们选择单词0 ^p 1 ^p 2 ^p作为s。由于s是A ₁的元素并且s比p长，泵引理确保s可以分为三部分，s = xyz，其中对于每个i ≥ 0，单词xy^和z属于A ₁。让我们考虑两种可能性：

1.29 (a) Assume that the language A ₁ = { 0 ⁿ 1 ⁿ 2 ⁿ : n ≥ 0} is regular. Let p be the pumping length that follows from the Pumping Lemma. We choose s to be the word 0 ^p 1 ^p 2 ^p . Since s is an element of A ₁ and s is longer than p , the Pumping Lemma ensures that s can be partitioned into three parts, s = xyz , where for each i ≥ 0 the word xy ^and z belong to A ₁ . Consider two possibilities:

1. 单词y由全0、全1或全2组成。在这些情况下，单词xyyz将不具有相同数量的数字0、1和2。因此，xyyz不属于语言A ₁，这意味着矛盾。

1. The word y consists of all 0s , all 1s , or all 2s . In these cases the word xyyz will not have an equal number of digits 0 , 1 , and 2. Thus xyyz does not belong to the language A ₁ , which means a contradiction.

2. 单词y包含不止一种类型的符号。在这种情况下，xyyz将包含乱序的0、1或2 。因此，xyyz不属于语言A ₁，这是矛盾的。

2. The word y contains more than one kind of symbol. In this case xyyz will contain 0 , 1 or 2 out of order. Thus xyyz does not belong to the language A ₁ , a contradiction.

在每种情况下我们都得到了一个矛盾。因此，A ₁不是正则语言。

In each case we have obtained a contradiction. Therefore A ₁ is not a regular language.

(c)假设A ₃ = { a ^{2 n} : n ≥ 0} 是正则的。令p为由泵浦引理得出的泵浦长度。我们选择a ^{2 p}作为s词。由于s是A ₃的元素并且s比p长，泵引理确保s可以分为三部分，s = xyz，其中对于每个i ≥ 0，单词xy^和z属于A ₃。

(c) Suppose A ₃ = { a ^{2 n} : n ≥ 0} is regular. Let p be the pumping length that follows from the Pumping Lemma. We choose as s the word a ^{2 p} . Since s is an element of A ₃ and s is longer than p , the Pumping Lemma ensures that s can be partitioned into three parts, s = xyz , where for each i ≥ 0 the word xy ^and z belong to A ₃ .

引理的第三个条件说 | XY | ≤ p。此外，p < 2 ^p且 | y | < 2 ^p。因此| xyyz | = | XYZ |+| y | < ^2p + 2p ⁼ 2p ^+1。第二个条件要求 | y | > 0，因此 2 ^p < | xyyz | < ^{2p +1}。因此xyyz的长度不能是 2 的幂。因此xyyz不是A ₃的元素，这是一个矛盾。因此，A ₃不是正则语言。

The third condition of the lemma says that | xy | ≤ p . Additionally , p < 2 ^p and therefore | y | < 2 ^p . Hence, | xyyz | = | xyz |+| y | < 2 ^p +2 ^p = 2 ^{p +1} . The second condition requires that | y | > 0, hence 2 ^p < | xyyz | < 2 ^{p +1} . The length of xyyz cannot therefore be a power of 2. Hence xyyz is not an element of A ₃ , which is a contradiction. Hence, A ₃ is not a regular language.

1.40 (a)设M = ( Q , Σ , δ , q ₀ , F ) 为识别A的 DFA ，其中A是某种正则语言。我们构造M' = ( Q' , Σ , δ' , q ₀ ' , F' ) 来识别NOPREFIX语言(A)，如下所示：

1.40 (a) Let M = ( Q , Σ , δ , q ₀ , F ) be a DFA recognizing A , where A is some regular language. We construct M' = ( Q' , Σ , δ' , q ₀ ' , F' ) recognizing the NOPREFIX language (A) as follows:

1. Q' = Q。

1. Q' = Q .

2. 对于r ∈ Q'和a ∈ Σ，我们定义δ' ( r , a ) =

2. For r ∈ Q' and a ∈ Σ , we define δ' ( r , a ) =

3. q ₀ ' = q ₀。

3. q ₀ ' = q ₀ .

4. F' = F。

4. F' = F .

1.44 设M _B = ( Q _B , Σ , δ _B , q _B , F _B ) 和M _C = ( Q _C , Σ , δ _C , q _C , F _C ) 分别为识别语言B和C的 DFA 。我们构造一个 NFA M = ( Q , Σ , δ , q ₀ , F ) 来识别B Ø ¹ C如下。为了确定输入数据是否属于B ‐ ¹ C，机器M检查w ∈ B并并行非确定性地猜测单词y ，该单词包含与w中相同数量的 1并检查y ∈ C 。

1.44 Let M _B = ( Q _B , Σ , δ _B , q _B , F _B ) and M _C = ( Q _C , Σ , δ _C , q _C , F _C ) be DFAs recognizing languages ​​B and C , respectively . We construct an NFA M = ( Q , Σ , δ , q ₀ , F ) recognizing B ¬ ¹ C as follows. To determine whether the input belongs to B ¬ ¹ C , the machine M checks that w ∈ B and in parallel nondeterministically guesses a word y that contains the same number of 1s as contained in w and checks that y ∈ C .

1. Q = Q _B × Q _C。

1. Q = QB × QC ._​​

2. 对于 ( q , r ) ε Q和a ε Σ _ε，我们定义

2. For ( q , r ) ∈ Q and a ∈ Σ _ε , we define

3. q ₀ = ( q _B , q _C )。

3. q ₀ = ( q _B , q _C ).

4. F = F _B × F _C。

4. F = F _B × F _C .

1.46 (b)令B = { 0 ^m 1 ⁿ : m ≠ n }。我们注意到B ∩ 0 ^* 1 ^* = { 0 ^k 1 ^k : k ≥ 0}。如果B是正则语言，那么B就是正则语言，这同样适用于B ∩ 0 ^* 1 ^*。然而，我们已经知道 { 0 ^k 1 ^k : k ≥ 0} 不是正则语言，因此B不可能是正则语言。

1.46 (b) Let B = { 0 ^m 1 ⁿ : m ≠ n }. We note that B ∩ 0 ^* 1 ^* = { 0 ^k 1 ^k : k ≥ 0}. If B were a regular language, then B would be regular and the same would be true for B ∩ 0 ^* 1 ^* . However, we already know that { 0 ^k 1 ^k : k ≥ 0} is not a regular language, so B cannot be regular.

或者，我们可以直接使用泵引理来证明B的非正则性，尽管这更困难。假设B = { 0 ^m 1 ⁿ : m  ¹  n } 是正则语言。令p为由泵浦引理得出的泵浦长度。请注意，p！可被 1 到p 的所有整数整除，其中p ! = p ( p − 1)( p − 2)…1。字s = 0 ^p 1 ^{p + p !} ε B i | s | ≥ p。因此，根据泵送引理，s可以表示为xyz，其中x = 0 ^a、y = 0 ^b且z = 0 ^c 1 ^{p + p！}，其中b ≥ 1 且a + b + c = p。令s'为单词xy^和+1 z，其中i = p !/ b。那么y ⁱ = 0 ^p！，所以y ^{i +1} = 0 ^{b + p！}类似地s' = 0 ^{a + b + c + p！}1p ^{+ p！​}。这给出s' = 0 ^{p + p！}1 ^{p + p !} ∉B——矛盾。

Alternatively, we can prove the irregularity of B directly using the Pumping Lemma, although this procedure is more difficult. Suppose B = { 0 ^m 1 ⁿ : m  ¹  n } is a regular language. Let p be the pumping length, which follows from the Pumping Lemma. Note that p ! is divisible by all integers from 1 to p , where p ! = p ( p − 1)( p − 2)…1. The word s = 0 ^p 1 ^{p + p !} ∈ B and | s | ≥ p . Therefore, by the Pumping Lemma, s can be represented as xyz , where x = 0 ^a , y = 0 ^b , and z = 0 ^c 1 ^{p + p !} , where b ≥ 1 and a + b + c = p . Let s' be a word xy ^{i +1} z , where i = p !/ b . Then y ⁱ = 0 ^{p !} , therefore y ^{i +1} = 0 ^{b + p !} and analogously s' = 0 ^{a + b + c + p !} 1 ^{p + p !} . This gives s' = 0 ^{p + p !} 1 ^{p + p !} ∉ B – a contradiction.

1.50 相反，假设某个 FST T对于输入w返回w R。考虑输入字00和01。对于输入00 T必须返回00，对于输入01 T必须返回10。在这两种情况下，第一个输入位都是0，但第一个输出位不同。这种类型的行为对于 FST 来说是不可能的，因为它在读取第二个输入符号之前创建第一个输出符号。因此，这样的FST不可能存在。

1.50 Suppose, on the contrary, that a certain FST T returns w R for an input w . Consider the input words 00 and 01 . For an input of 00, T must return 00 , and for an input of 01 , T must return 10 . In both cases, the first input bit is 0 , but the first output bits are different. Such an operation is impossible for an FST, since it produces its first output symbol before reading the second input symbol. Hence, such an FST cannot exist.

1.52 (a)我们将间接证明这个定理。令M为识别L 的k状态 DFA 。与定理相反，我们假设L的索引大于k。这意味着某些具有超过k 个元素的X可以通过L成对区分。由于M具有k 个状态，鸽笼规则意味着X包含两个不同的词x和y，使得δ ( q ₀ , x ) = δ ( q ₀ , y ) 。回想一下，δ ( q ₀ , x ) 是M在初始状态q ₀开始并读取输入字x后所处的状态。那么对于任何单词z ∈ Σ ^*，δ ( q ₀ , xz ) = δ ( q ₀ , yz ) 。因此，要么xz和yz都属于L，要么都不属于L。然而，那么x和y不能被L区分，这与X可以被L成对区分的假设相矛盾。

1.52 (a) We prove this theorem indirectly. Let M be a k -state DFA that recognizes L . Suppose, conversely, that L has index greater than k . This means that some X with more than k elements is pairwise distinguishable by L . Since M has k states, the pigeonhole principle implies that X contains two distinct words x and y , such that δ ( q ₀ , x ) = δ ( q ₀ , y ). Recall that δ ( q ₀ , x ) is the state that M is in after starting in the initial state q ₀ and reading the input word x . Then for any word z ∈ Σ ^* , δ ( q ₀ , xz ) = δ ( q ₀ , yz ). Thus, either both xz and yz belong to L , or neither of them belongs to L . However, then x and y are not discriminable by L , which contradicts the assumption that X is pairwise discriminable by L .

(b)令X = { s ₁ , …, s _k } 为可通过L区分的成对集合。我们将构造一个具有k 个状态的 DFA M = ( Q , Σ , δ , q ₀ , F ) ，识别L。令Q = { q ₁ , …, q _k };我们将δ ( q _i , a ) 定义为q _j，其中s _j == _L s _i a（关系== _L在问题 1.51 中定义）。请注意，对于某些s _j ∈ X ， s _j == _L s _i a；否则，集合X ∪ s _i a将具有k + 1 个元素，并且可以通过L成对区分，这与L具有索引k的假设相矛盾。令F = { q _i : s _i ∈ L }。令初始状态q ₀为q _i ，使得s _i ≠ _L ε。M的构造使得对于任何状态q _i， { s : δ ( q ₀ , s ) = q _i } = { s : s ≡ _L s _i }。因此，M识别语言L。

(b) Let X = { s ₁ , …, s _k } be a pairwise distinguishable set by L . We construct a DFA M = ( Q , Σ , δ , q ₀ , F ) with k states, recognizing L . Let Q = { q ₁ , …, q _k }; we define δ ( q _i , a ) to be q _j , where s _j ≡ _L s _i a (the relation ≡ _L was defined in Problem 1.51). Note that s _j ≡ _L s _i a for some s _j ∈ X ; otherwise the set X ∪ s _i a would have k + 1 elements and would be pairwise distinguishable by L , which would contradict the assumption that L has index k . Let F = { q _i : s _i ∈ L }. Let the initial state q ₀ be such q _i that s _i ≡ _L ε . M is constructed in such a way that for any state q _i , { s : δ ( q ₀ , s ) = q _i } = { s : s ≡ _L s _i }. Hence M recognizes the language L .

(c)假设L是正则语言，令k为识别L的DFA的状态数。然后，基于点(a)，L的索引至多等于k。相反，如果L具有索引k，则基于点 ( b )，它可以被k 个状态的 DFA 识别，因此是正则的。为了表明索引L是接受它的最小 DFA 的量值，假设索引L恰好是k。那么，由 ( b ) 可知，存在一个接受L 的k状态 DFA 。这是最小的此类 DFA，因为如果存在更小的 DFA，我们可以从 (a) 证明索引L小于k。

(c) Suppose L is a regular language and let k be the number of states of a DFA that recognizes L . Then by (a) L has index at most k . Conversely, if L has index k , then by ( b ) it is recognized by a k - state DFA and is therefore regular. To show that the index of L is the size of the smallest DFA that accepts it, suppose that the index of L is exactly k . Then by ( b ) there exists a k -state DFA that accepts L . This is the smallest such DFA, since if there were a smaller one, we could show by (a) that the index of L is less than k .

1.55 (a)最小泵长度为4。字000属于该语言但不能泵，因此3不是该语言的泵长度。如果s 的长度为 4 或更大，则它包含 1s。通过将单词s分为xyz，其中x是000，y是素数 1 ，以及其他所有内容，我们满足泵引理的所有三个条件。

1.55 (a) The minimum pumping length is 4. The word 000 belongs to the language but cannot be pumped, so 3 is not a pumping length for that language. If s has length 4 or greater, then it contains ones. By splitting the word s into xyz , where x is 000 , y is the first one, and everything else, we satisfy all three conditions of the Pumping Lemma.

(b)最小泵送长度为1。泵送长度不能为0，因为单词ε属于该语言但不能被泵送。该语言中的任何非空单词都可以拆分为xyz，其中x、y和z分别是ε、第一个字符和余数。该除法满足引理的三个条件。

(b) The minimal pumping length is 1. The pumping length cannot be 0, since the word ε belongs to the language but cannot be pumped. Every nonempty word in the language can be partitioned into xyz , where x , y , and z are ε , the first character, and the remainder, respectively . This partitioning satisfies the three conditions of the lemma.

(d)最小泵长度为 3。泵长度不能为 2，因为单词11属于该语言并且不能被泵。令s为语言中长度至少为 3 的单词。如果s由0 ^* 1 ⁺ 0 ⁺ 1 ^*生成，并且从0或11开始，那么我们可以写s = xyz，其中x = ε，y是第一个符号，z是s的剩余部分。如果s由0 ^* 1 ⁺ 0 ⁺ 1 ^*生成并从10开始，那么我们写s = xyz，其中x = 10，y是下一个符号，余数是s。以这种方式分裂s表明它可以被泵送。如果s是由10 ^* 1生成的，那么我们可以将其写为xyz，其中x = 1，y = 0，z是s的余数。该部门还提供泵送s的能力。

(d) The minimum pumping length is 3. The pumping length cannot be 2, since word 11 belongs to the language and cannot be pumped. Let s be a word from the language of length at least 3. If s is generated by 0 ^* 1 ⁺ 0 ⁺ 1 ^* and starts with either 0 or 11 , then we can write s = xyz , where x = ε , y is the first symbol, and z is the remainder of s . If s is generated by 0 ^* 1 ⁺ 0 ⁺ 1 ^* and starts with 10 , then we write s = xyz , where x = 10 , y is the next symbol, and z is the remainder of s . Partitioning s in this way shows that it can be pumped. If s is generated by 10 ^* 1 , then we can write it as xyz , where x = 1 , y = 0 , and z is the remainder of s . This partitioning also provides the possibility of pumping s .